Fundamentals of Thermodynamics 9th editionch15

CHAPTER 15 SOLUTION

Borgnakke and Sonntag

CONTENT SUBSECTION

PROB NO.

In-Text concept questions Concept-Study Guide Problems Fuels and the Combustion Process Energy Equation, Enthalpy of Formation Enthalpy of Combustion and heating Value Adiabatic Flame Temperature Second Law for the Combustion Process Problems Involving Generalized Charts or Real Mixtures Fuel cells Combustion Applications and Efficiency Review Problems

a-k 1-18 19-33 34-49 50-71

104-109 110-119 120-130 131-148

English Unit Problems

149-182

72-90 91-103

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In-Text Concept Questions



15.a

How many kmol of air are needed to burn 1 kmol of carbon? Combustion Eq.:

C + O2

⇒

1 CO2

One kmol of O2 is required to form one kmol CO 2 . Since air is 21 % O2 , this means 4.76 kmol of air.



15.b

If I burn 1 kmol of hydrogen H2 with 6 kmol of air what is the air-fuel ratio on a mole basis and what is the percent theoretical air? Combustion Eq. stoichiometric: H2 + νO2(O2 + 3.76 N2)

νO2 = 0.5 ; Six kmol of air is:

⇒

1 H2O + 3.76 νO2 N2

(A/F)S = νO2 × (1 + 3.76) / 1 = 2.38 1.26 O + 4.74 N . 2

2

The A/F mole ratio is 6, so the percent theoretical air is (A/F)ac 6 %Theoretical air = × 100 = 2.38 × 100 = 252 % (A/F)S



15.c

For the 110% theoretical air in Eq.15.8 what is the equivalence ratio? Is that mixture rich or lean? 110% Theoretical air means also

AF = 1.1 AFS

so from the definition in Eq.15.6 AFS 1 Φ = AF = 1.10 = 0.909

a lean mixture

15.d

In most cases combustion products are exhausted above the dew point. Why? If any water in the products condenses, it will be acidic due to the other gases in the products. There are always minute amounts of unburned or partially burned fuel and intermediate species in the products that can combine with water and create a very corrosive mixture.



15.e

How is a fuel enthalpy of combustion connected to its enthalpy of formation? The enthalpy of combustion of a fuel is the difference in enthalpy of the products and reactants for the combustion involving the fuel, these enthalpies include the various enthalpies of formation. At reference condition:

° ° ° ° -0 HRP = HP - HR = HP - hf fuel vap or liq

15.f

What is the higher and lower heating value HHV, LHV of n-butane? The heating value is the negative of the enthalpy of combustion. From Table 15.3, the HHV of gaseous n-butane at 25 °C is HHV = 49 500 kJ/kg, and the corresponding LHV is LHV = 45 714 kJ/kg. Notice the table is on a mass basis (per kg fuel).



15.g

What is the value of hfg for n-octane? This can be obtained from two places. From Table A.10 we get -0 -0 hfg = (hf vap– hf liq) / M = [(–208 600 – (–250 105)] / 114.232 = 363 kJ/kg The hfg of a fuel listed in Table 15.3 is the difference between the first two columns in the table (or the third and fourth). For n-octane, this value is hfg = –47 893 – (–48 256) = 363 kJ/kg To see this remember ° ° ° ° -0 HRP = HP - HR = HP - hf fuel vap or liq

so when we take the difference between fuel as gas and fuel as liquid all other terms will cancel out leaving hfg for the fuel.



15.h

What happens to the adiabatic flame temperature if I burn rich or lean? The higher the percent theoretical air used in combustion (the leaner), the larger the number of moles of products, especially nitrogen, which decreases the adiabatic flame temperature. Burning rich causes incomplete combustion, however, with a smaller release of energy. Experimentally the highest temperature is reached for slightly rich.

Heavy molecules show up as yellow. Oxygen diffuses in from the air and the fuel evaporates from the wick. As air mixes in, the flame cools.

15.i

Is the irreversibility in a combustion process significant? Why is that? A combustion process is highly irreversible with a large increase in entropy. It takes place at a rapid rate, due to large driving forces, and results in stable products of combustion that have little or no tendency to return to their former constituents and states.



15.j

If the air-fuel ratio is larger than stoichiometric is it more or less reversible? Less reversible more irreversible. The excess oxidizer (air) is being heated up, Q over a finite temperature difference is an irreversible process. The same is true for A/F smaller than one where the excess fuel is heated up. 15.k

What makes the fuel cell attractive from a power-generating point of view? Fuel cells are attractive for power generation because their direct output is electrical energy. They also have a much higher power density as power per unit volume or power per unit mass and thus can be used in mobile applications.



Concept-Study Guide Problems



15.1

Is mass conserved in combustion? Is the number of moles constant? Yes mass is conserved as we do not consider the quantum physic effect of mass and energy conversion. Moles are not conserved, what is conserved are the atoms so each atom in the reactants must be found in the products. 15.2

Does all combustion take place with air? Most combustion takes place with air as the oxidizer, special cases do occur where a different oxidizer is used. To reach very high temperatures a gas welder uses a bottle of pure oxygen (and one with the fuel) instead of air and then the flame can cut through steel.



15.3

Why would I sometimes need A/F on a mole basis? on a mass basis? If you want to meter (measure) the fuel and air flows it can be done as a . .− volume flowrate which is proportional to moles (PV = nRT) in which case concentrations on a mole basis are needed. The fuel and air flows can also be measured with a method that measures . mass flow rate m or if you are filling up tanks to store the fuel and oxidizer as in a rocket in both cases the concentrations on a mass basis are needed.

15.4

Why is there no significant difference between the number of moles of reactants versus products in combustion of hydrocarbon fuels with air? In most hydrocarbon fuels, there are approximately twice as many hydrogen atoms as carbon atoms, so the numbers of moles of CO2 and H2O in the products are roughly equal, the total of which is not too different from the O2 required in the reactants. The number of excess O2 is the same in reactants and products. The total number of moles is dominated by the N2 in each, especially with excess air.



15.5

Why are products measured on a dry basis? Combustion products have traditionally been measured by passing the gas mixture through a series of solutions that selectively absorb the components oneby-one and measuring the resulting gas volume decreases. The water component is condensed out in these processes, leaving the others – that is, a dry basis. Other and newer instruments measure the concentrations by optical means and these are sensitive to moisture content, which can corrode the surfaces and destroy the sensors. If the water stays in the mixture it typically has to stay hot to prevent condensation at undesirable locations where that would alter the concentrations of the remaining gas mixture components.



15.6

What is the dew point of hydrogen burned with stoichiometric pure oxygen? air? For H2 burned with pure oxygen H2 + 0.5 O2

⇒

1 H 2O

with the only product as water, so the dew-point at 100 kPa pressure is 99.6 C. For H2 burned with stoichiometric air H2 + 0.5 (O2 + 3.76 N2)

⇒

1 H2O + 1.88 N2

the product mixture is water and nitrogen. The partial pressure of the water at a total pressure of 100 kPa is 1 Pv = Ptot yv = 100 × = 34.7 kPa, 1 + 1.88 corresponding to a dew-point of 72.3 C .



15.7

How does the dew point change as equivalence ratio goes from 0.9 to 1 to 1.1? For a given amount of water in the products, the smaller the total number of moles of other gases is (as Φ increases), the higher the partial pressure of the water and therefore the dew-point temperature. As Φ becomes greater than 1.0, there will be incomplete combustion, and the resulting composition will be affected to have some unburned fuel and therefore relative less water. The relative maximum amount of water is then at a stoichiometric mixture Φ = 1 and this is also maximum dew point temperature. 15.8

Why does combustion contribute to global warming? Any combustion of a hydrocarbon fuel produces carbon dioxide, which in the atmosphere is a contributing factor to global warming. Carbon dioxide absorbs radiation over a wide spectrum and thus heats the atmosphere. This is not just man-made, but nature has forest fires and volcanic action that liberate gases into the atmosphere.



15.9

What is the enthalpy of formation for oxygen as O2? If O? For CO2? The enthalpy of formation of O2 is zero, by choice of the reference base. Relative to this base, the value for the monatomic form O is h°f O = +249 170 kJ/kmol (Table A.9), and the value for CO2 is h°f CO2 = –393 522 kJ/kmol (Table A.9 or A.10).



15.10

If the nitrogen content of air could be lowered will the adiabatic flame temperature increase or decrease? Since that would reduce the number of nitrogen molecules in the products it will increase the adiabatic flame temperature, more energy per molecule. Energy Eq.:

∆HP = HV + ∆HR = given ∆HP = νCO ∆hCO + νH O ∆hH 2

2

2

O 2

-

+ νN

∆hN 2

at Tad 2

15.11

Does the enthalpy of combustion depend on AF? NO. HRP = HP - HR depends on T, P in general. Assuming ideal gas it does not depend on P. It is assumed the number is scaled to 1 kmol or 1 kg of fuel burned.



15.12

Why do some fuels not have entries for liquid fuel in Table 15.3? Those fuels cannot exist as liquids at 25 °C (above their critical temperature).



15.13

Is a heating value a fixed number for a fuel? Yes. The heating value is the negative of the enthalpy of combustion and is only useful if it is a well defined number. It is fixed, but there are normally 4 values quoted as the higher (water liquid) and lower (water vapor) heating value for the fuel as a liquid or vapor if applicable. It should also be clear whether it is the enthalpies or the internal energies that are used, i.e. do you have a flow or a non flow situation.

So these qualifiers must accompany the number for it to be unique.



15.14

Is an adiabatic flame temperature a fixed number for a fuel? Qualified Yes. It is a single number if we assume reactants are supplied at the reference conditions in a stoichiometric ratio and we know the phase of the fuel (liquid or gas) and the type of oxidizer (air or pure oxygen). Also you need to know if the combustion takes place at constant pressure (the most common) or it may take place at constant volume (unusual). If anyone of these conditions is changed the adiabatic flame temperature is changed. This is less fixed than the heating value, more variables can change the adiabatic flame temperature, say, adding an inert gas will lower the flame temperature, but keep the heating value the same. Chemical equilibrium reactions and incomplete combustion makes the real flame temperature lower than the theoretical adiabatic flame temperature.



15.15

Does it make a difference for the enthalpy of combustion whether I burn with pure oxygen or air? What about the adiabatic flame temperature? No difference in the enthalpy of combustion – the nitrogen in the air is the same in the reactants and products, and its enthalpy cancels out. The adiabatic flame temperature is much lower for combustion with air, because a significant part of the energy release from the combustion goes into heating the nitrogen (as well as the other products) to the flame temperature.



15.16

A welder uses a bottle with acetylene and a bottle with oxygen. Why does he use oxygen instead of air?

The oxygen in the air comes with nitrogen that is also being heated up to the flame temperature. The heating value is being shared among all the product components so when they increase (having the nitrogen also) the adiabatic flame temperature drops correspondingly.



15.17

Some gas-welding is done using bottles of fuel, oxygen and argon. Why do you think argon is used? Argon is used to shield the welding spot from the oxygen in the surrounding air. If oxygen is on the surface of the liquid metal it may form metal-oxides creating a weak location in the weld. 15.18

Is combustion a reversible process? NO. It is a highly irreversible process.



Fuels and the Combustion Process



15.19

In a picnic grill gaseous propane is fed to a burner together with stoichiometric air. Find the air-fuel ratio on a mass basis and the total amount of reactant mass for 1 kg propane burned. The reaction equation for stoichiometric mixture is: C3H8 + νO2 (O2 + 3.76 N2)

→

a H2O + b CO2 + c N2

C balance: 3 = b ; H balance: 8 = 2a

=>

a=4

2 νO2 = a + 2b = 4 + 2 × 3

O balance: N2 balance:

=>

νO2 = 5

3.76 νO2 = c = 18.8

The reaction is C3H8 + 5 (O2 + 3.76 N2) AFmol = AFmass = AFmol

→ 4 H2O + 3 CO2 + 18.8 N2

5 + 18.8 = 23.8 1 Mair Mfuel

= 23.8 ×

28.97 = 15.64 44.097

Total mass of reactants = 1 + 15.64 = 16.64 kg



15.20

Calculate the theoretical air-fuel ratio on a mass and mole basis for the combustion of ethanol, C2H5OH. Reaction Eq.: C2H5OH +

νO2(O2 + 3.76N2) ⇒

aCO2 + bH2O + cN2

Do the atom balance Balance C: 2 = a Balance H: 6 = 2b

⇒

b=3

Balance O: 1 + 2νO2 = 2a + b = 4 + 3 = 7

⇒ νO2 = 3

(A/F)mol = νO2(1 + 3.76)/1 = 3 × 4.76 = 14.28 (A/F)mass = (νO2MO2 + νN2 MN2)/MFuel = (3×31.999 + 11.28×28.013) / 46.069 = 8.943



15.21

A certain fuel oil has the composition C10H22. If this fuel is burned with 150% theoretical air, what is the composition of the products of combustion?

C10H22 + (1/φ) νO

2

(O2 + 3.76 N2) → a H2O + b CO2 + c N2 + d O2

Stoichiometric combustion:

φ = 1,

d = 0,

C balance: b = 10 H balance: a = 22/2 = 11, O balance: 2 νO = a + 2b = 11 + 20 = 31 2

Actual case:

1/φ = 1.5 =>

H balance: a = 11,

=>

νO2 = 15.5

νO2 = 1.5 × 15.5 = 23.25 C balance:

b = 10,

N balance: c = 23.25 × 3.76 = 87.42 O2 balance:

d = 23.25 - 10 - 11/2 = 7.75 (excess oxygen)

The combustion process is C10H22 + 23.25 (O2 + 3.76 N2)

→ 11 H2O + 10 CO2 + 87.42 N2 + 7.75 O2



15.22

Methane is burned with 200% theoretical air. Find the composition and the dew point of the products. The reaction equation for stoichiometric mixture is: CH4 + νO2 (O2 + 3.76 N2)

→


C balance:

1=b;

O balance:

2 νO2 = a + 2b = 2 + 2 × 1

N2 balance:

H balance:

νO2 = 2

3.76 νO2 = c = 7.52

200% theoretical air:

νO2 = 2 × 2 = 4

CH4 + νO2 (O2 + 3.76 N2) N2 balance:

=>

4 = 2a

→

so now more O2 and N2

a H2O + b CO2 + c N2 + d O2

3.76 νO2 = c = 15.04

Extra oxygen: d = 4 – 1 - 1 = 2 Products:

2 H2O + 1 CO2 + 15.04 N2 + 2 O2 2 = 0.0998 1 + 2 + 2 + 15.04

Water vapor mole fraction:

yv =

Partial water vapor pressure:

Pv = yv Po = 0.0998 × 101 = 9.98 kPa

Pg(Tdew) = Pv = 9.98 kPa

⇒

Tdew = 45.8oC



15.23

Natural gas B from Table 15.2 is burned with 20% excess air. Determine the composition of the products.

The reaction equation (stoichiometric and complete combustion) with the fuel composition is: 60.1 CH4 + 14.8 C2H6 + 13.4 C3H8 + 4.2 C4H10 + 7.5 N2 + νO2 (O2 + 3.76 N2)


→

C balance:

60.1 + 2×14.8 + 3×13.4 + 4×4.2 = b = 146.7

H balance:

4×60.1 + 6×14.8 + 8×13.4 + 10×4.2 = 2a = 478.4

O balance:

2 νO2 = a + 2b = 239.2 + 2×146.7

N2 balance:

a = 239.2

7.5 + 3.76 νO2 = c = 1008.8

20% excess air:

νO2 = 1.2×266.3 = 319.56

Extra oxygen: d = 319.56 - 266.3 = 53.26, Products:

⇒ ⇒ νO2 = 266.3

so now more O2 and N2 c = 7.5 + 3.76×319.56 = 1209

239.2 H2O + 146.7 CO2 + 1209 N2 + 53.26 O2

To the expert the color of the flame can tell about the composition. It can also tell about other gases present if they have distinct color emission.



15.24

For complete stoichiometric combustion of gasoline, C7H17, determine the fuel molecular weight, the combustion products, and the mass of carbon dioxide produced per kg of fuel burned. Stoichiometric combustion: C7H17 + νO

2

(O2 + 3.76 N2) →


C balance: H balance:

7=b 17 = 2a,

O balance:

2 νO = a + 2b = 8.5 + 14 = 22.5

N balance:

c = 3.76

⇒

2

νO

2

a = 8.5

⇒ νO

2

= 11.25

= 3.76 × 11.25 = 42.3

MFUEL = 7 MC + 17 MH = 7 × 12.011 + 17 × 1.008 = 101.213 mCO2 7 MCO2 7 × 44.01 = = = 3.044 kg CO2 per kg fuel mFUEL MFUEL 101.213



15.25

A Pennsylvania coal contains 74.2% C, 5.1% H, 6.7% O, (dry basis, mass percent) plus ash and small percentages of N and S. This coal is fed into a gasifier along with oxygen and steam, as shown in Fig. P14.26. The exiting product gas composition is measured on a mole basis to: 39.9% CO, 30.8% H 2, 11.4% CO2, 16.4% H2O plus small percentages of CH4, N2, and H2S. How many kilograms of coal are required to produce 100 kmol of product gas? How much oxygen and steam are required? Convert the mass concentrations to number of kmol per 100 kg coal: C : n = 74.2/12.01 = 6.178 H2: n = 5.1/2.016 = 2.530 O2: n = 6.7/31.999 = 0.209 Now the combustion equation reads x(6.178 C + 2.53 H2 + 0.209 O2) + y H2O + z O2 in and 39.9 CO + 30.8 H2 + 11.4 CO2 + 16.4 H2O out in 100 kmol of mix out Now we can do the atom balance to find (x, y, z)

→

C balance:

6.178 x = 39.9 + 11.4

H2 balance:

2.53×8.304 + y = 30.8 + 16.4

O2 balance: 0.209 × 8.304 +

x = 8.304

→

y = 26.191

26.191 39.9 16.4 +z= + 11.4 + 2 2 2

→ z = 24.719

Therefore, for 100 kmol of mixture out require:

830.4 kg of coal 26.191 kmol of steam 24.719 kmol of oxygen



15.26

Liquid propane is burned with dry air. A volumetric analysis of the products of combustion yields the following volume percent composition on a dry basis: 8.6% CO2, 0.6% CO, 7.2% O2 and 83.6% N2. Determine the percent of theoretical air used in this combustion process. a C3H8 + b O2 + c N2

→ 8.6 CO2 + 0.6 CO + d H2O + 7.2 O2 + 83.6 N2 ⇒ a = 3.067

C balance:

3a = 8.6 + 0.6 = 9.2

H2 balance: N2 balance:

4a = d ⇒ d = 12.267 c = 83.6

O2 balance:

b = 8.6 +

Air-Fuel ratio =

0.6 12.267 + + 7.2 = 22.234 2 2

22.234 + 83.6 = 34.51 3.067

Theoretical: C3H8 + 5 O2 + 18.8 N2

→

⇒ theo. A-F ratio = % theoretical air =

3 CO2 + 4 H2O + 18.8 N2 5 + 18.8 = 23.8 1

34.51 × 100 % = 145 % 23.8



15.27

In a combustion process with decane, C 10H22, and air, the dry product mole fractions are 83.61% N2, 4.91% O2, 10.56% CO2 and 0.92% CO. Find the equivalence ratio and the percent theoretical air of the reactants.

x C10H22 + (1/φ) νO

2

(O2 + 3.76 N2) → a H2O + b CO2 + c CO + d N2 + e O2

Stoichiometric combustion:

φ = 1,

c = 0, e = 0,

C balance:

b = 10x

H balance:

a = 22x/2 = 11x,

O balance:

2 νO = a + 2b = 11x + 20x = 31x

νO

2

= 15.5x,

2

νN

2

Actual combustion:

⇒

= 58.28x d = 83.61

(A/F)s = (νO

2

+

νN

)/x = 73.78

2

→

N balance:

(1/φ) νO × 3.76 = 83.61

C balance:

10x = 10.56 + 0.92 = 11.48

2

→ (1/φ) νO ⇒

2

= 22.24

x = 1.148

(A/F)ac = (1/φ) νO × 4.76/1.148 = 92.215 2

φ = (F/A)ac / (F/A)s = (A/F)s / (A/F)ac = 73.78 / 92.215 = 0.80

or

= 0.8

Percent theoretical air = 100 (1/φ) = 125%



15.28

A sample of pine bark has the following ultimate analysis on a dry basis, percent by mass: 5.6% H, 53.4% C, 0.1% S, 0.1% N, 37.9% O and 2.9% ash. This bark will be used as a fuel by burning it with 100% theoretical air in a furnace. Determine the air–fuel ratio on a mass basis. Converting the Bark Analysis from a mass basis: Substance

S

H2

C

O2

N2

c/M = kmol / 100 kg coal

0.1/32

5.6/2

53.4/12

37.9/32

0.1/28

0.003

2.80

4.45

1.184

0.004

Product

SO2

H2O

CO2

oxygen required

0.003

1.40

4.45

--

--

Combustion requires: 0.003 + 1.40 + 4.45 = 5.853 kmol O2 there is in the bark 1.184 kmol O2 so the net from air is 4.669 kmol O2 AF = (4.669 + 4.669 × 3.76) ×

28.97 kg air = 6.44 100 kg bark



15.29

Methanol, CH3OH, is burned with 200% theoretical air in an engine and the products are brought to 100 kPa, 30 °C. How much water is condensed per kilogram of fuel? CH3OH + νO {O2 + 3.76 N2} → CO2 + 2 H2O + 3.76 νO N2 2

Stoichiometric

2

νO2 S = 1.5 ⇒ νO2 AC = 3

Actual products: CO2 + 2 H2O + 1.5 O2 + 11.28 N2 Psat(30°C) = 4.246 kPa

⇒ yH2O = 0.04246 = 1 + ν ⇒ νH2O = 0.611 ⇒ MFu = 32.042

∆MH2O MFu

νH2O H2O

+ 1.5 + 11.28

∆νH2O cond = 2 - 0.611 = 1.389 =

kg H2O 1.389 × 18 = 0.781 32.042 kg fuel



15.30

The coal gasifier in an integrated gasification combined cycle (IGCC) power plant produces a gas mixture with the following volumetric percent composition: Product

CH4

H2

CO

CO2

N2

H 2O

H2S

NH3

% vol.

0.3

29.6

41.0

10.0

0.8

17.0

1.1

0.2

This gas is cooled to 40 °C, 3 MPa, and the H 2S and NH3 are removed in water scrubbers. Assuming that the resulting mixture, which is sent to the combustors, is saturated determine the mixture composition and the theoretical air–fuel ratio with in thewater, combustors. CH4

H2

CO

CO2

N2

n

0.3

29.6

41.0

10.0

0.8

81.7

yH2O =

nV nV+81.7

,

where

nV = number of moles of water vapor

Cool to 40°C PG = 7.384, P = 3000 kPa yH2O MAX =

nV 7.384 = 3000 nV+81.7

a) Mixture composition: CH4 H2

→

nV = 0.2016

CO

CO2

N2

H2O(v)

0.3 kmol 29.6 41.0 10.0 0.8 0.2016 81.9016 kmol (from 100 kmol of the srcinal gas mixture) 0.3 CH4 + 0.6 O2

→

0.3 CO2 + 0.6 H2O

29.6 H2 + 14.8 O2

→

29.6 H2O

41 CO + 20.5 O2

→

41 CO2

⇒ Number of moles of O2 = 0.6 + 14.8 + 20.5 = 35.9 Number of moles of air = 35.9 + 3.76 × 35.9 (N2) A/F =

28.97(35.9 + 3.76(35.9)) 0.3(16) + 29.6(2) + 41(28) + 10(44) + 0.8(28) + 0.2016(18)

= 2.95 kg air/kg fuel



15.31

Butane is burned with dry air at 40oC, 100 kPa with AF = 26 on a mass basis. For complete combustion find the equivalence ratio, % theoretical air and the dew point of the products. How much water (kg/kg fuel) is condensed out, if any, when the products are cooled down to ambient temperature? Solution: C4H10 + νO {O2 + 3.76 N2} → 4 CO2 + 5 H2O + 3.76 νO N2 2

2

νO2 S = 4 + 5/2 = 6.5;

Stoichiometric

3.76 νO2 = 24.44

(A/F)S = 6.5(31.999 + 3.76 × 28.013)/58.124 = 15.3574

νO

Actual:

2ac

=

(A/F)ac (A/F)s

νO

2S

=

26 6.5 = 11 15.3574

% Theoretical air =

26 100 = 169.3% 15.3574

Equivalence ratio

Φ = 1/1.693 = 0.59

Actual products:

4 CO2 + 5 H2O + 4.5 O2 + 41.36 N2

The water partial pressure becomes Pv = yv Ptot =

5 100 = 9.114 kPa 4 + 5 + 4.5 + 41.36

Tdew = 43.85oC Pg 40 = 7.348 kPa

⇒

yv max =

νH2O 7.384 = 100 4 + νH2O + 4.5 + 41.36

Solve for νH2O vap:

νH2O vap

= 3.975 still vapor,

νH2O LIQ = 5 – 3.975 = 1.025 mH2O LIQ mFuel

=

is liquid

1.025 × 18.015 = 0.318 58.124



15.32

The output gas mixture of a certain air–blown coal gasifier has the composition of producer gas as listed in Table 14.2. Consider the combustion of this gas with 120% theoretical air at 100 kPa pressure. Determine the dew point of the products and find how many kilograms of water will be condensed per kilogram of fuel if the products are cooled 10°C below the dew-point temperature.

{3 CH4 + 14 H2 + 50.9 N2 + 0.6 O2 + 27 CO + 4.5 CO2} + 31.1 O2 + 116.9 N2 → 34.5 CO2 + 20 H2O + 5.2 O2 + 167.8 N2 Products: yH2O = yH2O MAX = PG/100 =

⇒ PG = 8.79 kPa →

20 34.5 + 20 + 5.2 + 167.8

TDEW PT = 43.2°C

At T = 33.2°C, PG = 5.13 kPa yH2O =

nH2O 5.13 = 100 nH2O+34.5+5.2+167.8

→

nH2O = 11.22

8.78(18)

mH2O LIQ = 3(16) +14(2) +50.9(28) +0.6(32) +27(28) +4.5(44) = 0.0639 kg/kg fuel



15.33

The hot exhaust gas from an internal combustion engine is analyzed and found to have the following percent composition on a volumetric basis at the engine exhaust manifold. 10% CO2, 2% CO, 13% H2O, 3% O2 and 72% N2. This gas is fed to an exhaust gas reactor and mixed with a certain amount of air to eliminate the carbon monoxide, as shown in Fig. P15.33. It has been determined that a mole fraction of 10% oxygen in the mixture at state 3 will ensure that no CO remains. What must the ratio of flows be entering the reactor? Exhaust gas at state 1: CO 2 10 %, H2O 13%, CO 2%, O2 3%, N2 72%

1 Exh. gas

Exhaust gas at state 3:

2 Air

Reactor

CO = 0 %, O2 = 10 %

3 gas out

Reaction equation for the carbon monoxide

⇒ 0.02 CO + x O2 + 3.76x N2 → 0.02 CO2 + (x-0.01) O2 + 3.76x N2 At 3:

νCO2 = 0.10 + 0.02 = 0.12,

νH2O = 0.13

νO

νN

2

= (x-0.01) + 0.03 = x + 0.02

2

= 0.72 + 3.76x

or nTOT = 0.12 + 0.13 + x + 0.02 + 0.72 + 3.76x = 0.99 + 4.76x yO2 = 0.10 = or

x + 0.02 0.99 + 4.76x

→

x = 0.151

air 2 4.76x kmol air = = 0.718 Exh. Gas 1 1 kmol Exh. gas



Energy Equation, Enthalpy of Formation



15.34

Hydrogen is burned with stoichiometric air in a steady flow process where the reactants are supplied at 100 kPa, 298 K. The products are cooled to 800 K in a heat exchanger. Find the heat transfer per kmol hydrogen. The combustion equation is H2 + 0.5 × [ O2 + 3.76 N2 ]

→ 1 H2O + 1.88 N2

The reactants are at 298 K HR = 0 + 0 + 0 = 0 The products are at 800 K so water is gaseous and from A.9 HP = 1[-241 826 + 18 002)] +

1.88 (0 + 15 046)

= -195 538 kJ/kmol fuel

QCV = HP - HR = -195 538 kJ/kmol fuel



15.35

Butane gas and 200% theoretical air, both at 25 °C, enter a steady flow combustor. The products of combustion exits at 1000 K. Calculate the heat transfer from the combustor per kmol of butane burned. C4H10 + (1/φ) νO2(O2 + 3.76 N2)

→

a CO2 + b H2O + c N2 + d O2

First we need to find the stoichiometric air ( φ = 1, d = 0 ) C balance: 4 = a,

H balance: 10 = 2b =>

O balance: 2νO2 = 2a + b = 8 + 5 = 13 => Now we can do the actual air: (1/φ) = 2 =>

b=5

νO2 = 6.5 νO2 = 2 × 6.5 = 13

N balance: c = 3.76 νO2 = 48.88, O balance: d = 13 - 6.5 = 6.5 o

o

q = HR - HP = HR - HP - ∆HP

Energy Eq.: o

Table A.10: HR = -126 200 + 0 + 0 = -126 200 kJ/kmol fuel o

HP = 4 (-393 522) + 5(-241 826) + 0 + 0 = -2 783 218 kJ/kmol fuel The rest of the values are from Table A.9 at 1000 K

− − − − ∆hCO2 = 33397, ∆hN2 = 21463, ∆hO2 = 22703, ∆hH2O = 26000 kJ/kmol ∆H

= 4 × 33 397 + 5 × 26 000 + 48.88 × 21 463 + 6.5 × 22 703 P

= 1 460 269 kJ/kmol fuel

From the energy equation we get q = -126 200 –(-2 783 218) - 1 460 269 = 1 196 749 kJ/kmol butane



15.36

One alternative to using petroleum or natural gas as fuels is ethanol (C 2H5OH), which is commonly produced from grain by fermentation. Consider a combustion process in which liquid ethanol is burned with 120% theoretical air in a steady flow process. The reactants enter the combustion chamber at 25 °C, and the products exit at 60°C, 100 kPa. Calculate the heat transfer per kilomole of ethanol.

C2H5OH + 1.2 × 3 (O2 + 3.76 N2) → 2CO2 + 3H2O + 0.6O2 + 13.54N2 -0 Fuel: hf = -277 380 kJ/kmol for liquid from Table A.10, Products at 60°C, 100 kPa, check for condensation of water yH2O MIX =

nV MAX 19.94 = 100 nV MAX+2+0.6+13.54

=> nV MAX = 4.0 > 3

⇒

No liq.

HR = 1(-277 380) + 0 + 0 = -277 380 kJ/kmol fuel HP = 2(-393 522 + 1327) + 3(-241 826 + 1178) + 0.6(0 + 1032) + 13.54(0 + 1020) = -1 491 904 kJ/kmol fuel QCV = HP - HR = -1 214 524 kJ/kmol fuel



15.37

Do the previous problem with the ethanol fuel delivered as a vapor. One alternative to using petroleum or natural gas as fuels is ethanol (C 2H5OH), which is commonly produced from grain by fermentation. Consider a combustion process in which liquid ethanol is burned with 120% theoretical air in a steady flow process. The reactants enter the combustion chamber at 25 °C, and the products exit at 60°C, 100 kPa. Calculate the heat transfer per kilomole of ethanol. C2H5OH + 1.2 × 3 (O2 + 3.76 N2)

→

2CO2 + 3H2O + 0.6O2 + 13.54N2

-0 Fuel: hf = -235 000 kJ/kmol for IG from Table A.10 Products at 60°C, 100 kPa, check for condensation of water yH2O MIX =

nV MAX 19.94 = 100 nV MAX + 2 + 0.6 + 13.54

=> nV MAX = 4.0

The mixture can then hold 4 moles of vapor > 3 present

⇒

No liquid.

HR = 1(-235 000) + 0 + 0 = -235 000 kJ/kmol fuel HP = 2(-393 522 + 1327) + 3(-241 826 + 1178) + 0.6(0 + 1032) + 13.54(0 + 1020) = -1 491 904 kJ/kmol fuel QCV = HP - HR = -1 256 904 kJ/kmol fuel



15.38

As an alternative fuel consider liquid methanol burned with stoichiometric air both supplied at Po, To in a constant pressure process exhausting the products at 900 K. What is the heat transfer per kmol of fuel.

CH3OH + 1.5 (O2 + 3.76 N2)

→

1 CO2 + 2 H2O + 5.64 N2

o

Reactants at 25o C, products are at 900 K, 100 kPa. CH3OH: hf = –239 220 kJ/kmol from table A.10 for the liquid state HR = 1 hLIQ = –239 220 kJ/kmol fuel HP = 1(–393 522 + 28 030) + 2(–241 826 + 21 937) + 5.64(18 223) = –702 492 kJ/kmol fuel Q = HP - HR = –702 492 – (–239 220) = –463 272 kJ/kmol fuel



15.39

Another alternative fuel to be seriously considered is hydrogen. It can be produced from water by various techniques that are under extensive study. Its biggest problem at the present time are cost, storage, and safety. Repeat Problem 15.36 using hydrogen gas as the fuel instead of ethanol. H2 + 1.2 × 0.5 O2 + 1.2 × 3.76 × 0.5 N2

→ 1 H2O + 0.1 O2 + 2.256 N2

Products at 60°C, 100 kPa, check for condensation of water yH2O MAX =

nV MAX 19.94 = 100 nV MAX + 0.1 + 2.256

Solving, nV MAX = 0.587 < 1 => nV = 0.587, nLIQ = 0.413 HR = 0 + 0 + 0 = 0 Notice the products are at 60°C so add for water liquid from steam tables HP = 0.413[-285 830 + 18.015(251.1 - 104)] + 0.587(-241 826 + 1178) + 0.1(0 + 1032) + 2.256(0 + 1020) = -255 816 kJ QCV = HP - HR = -255 816 kJ/kmol fuel



15.40

The combustion of heptane C7H16 takes place in a steady flow burner where fuel and air are added as gases at P o, To. The mixture has 125% theoretical air and the products are going through a heat exchanger where they are cooled to 600 K.Find the heat transfer from the heat exchanger per kmol of heptane burned. The reaction equation for stoichiometric ratio is: C7H16 + vO2 (O2 + 3.76 N2)

=> 7CO2 + 8 H2O + vO2 × 3.76 N2

So the balance (C and H was done in equation) of oxygen gives v O2 = 7 + 4 = 11, and actual one is 11×1.25 = 13.75. Now the actual reaction equation is: C7H16 + 13.75 O2 + 51.7 N2

=> 7CO2 + 8 H2O + 51.7 N2+ 2.75 O2

To find the heat transfer take control volume as combustion chamber and heat exchanger HR + Q = HP

=>

o

o

Q = HP + ∆HP - HR

Take the enthalpies from Tables A.9 for the products and A.10 for the fuel Q = 7(-393 522 + 12 906) + 8 (-241 826 + 10 499) + 51.7(8894) + 2.75(9245) - (-187 900) = - 3 841 784 kJ/kmol fuel



15.41

In a new high-efficiency furnace, natural gas, assumed to be 90% methane and 10% ethane (by volume) and 110% theoretical air each enter at 25 °C, 100 kPa, and the products (assumed to be 100% gaseous) exit the furnace at 40 °C, 100 kPa. What is the heat transfer for this process? Compare this to an older furnace where the products exit at 250 °C, 100 kPa. 0.9 CH4 + 0.1 C 2 H 2 25 C, 100 kPa Products 40 C 100 kPa 110 % Air 25 C, 100 kPa

Heat transfer

0.9 CH4 + 0.1 C2H6 + 1.1 × 2.15 O2 + 3.76 × 2.365 N2

→ 1.1 CO2 + 2.1 H2O + 0.215 O2 + 8.892 N2 Fuel values from table A.10 and the rest from Table A.9 HR = 0.9(-74 873) + 0.1(-84 740) = -75860 kJ/kmol fuel H = 1.1(-393 522 + 562) + 2.1(-241 826 + 504) + 0.215(441) + 8.892(437) P

= -935 052 kJ/kmol fuel assuming all gas QCV = HP - HR = -859 192 kJ/kmol fuel b) TP = 250 oC HP = 1.1(-393 522 + 9346) + 2.1(-241 826 + 7740) + 0.215(6808) + 8.892(6597) = -854 050 kJ QCV = HP - HR = -778 190 kJ/kmol fuel



15.42

Repeat the previous problem, but take into account the actual phase behavior of the products exiting the furnace. 0.9 CH4 + 0.1 C2H6 + 1.1 × 2.15 O2 + 3.76 × 2.365 N2

→ 1.1 CO2 + 2.1 H2O + 0.215 O2 + 8.892 N2 Same as 14.49, except check products for saturation at 40 oC, 100 kPa yV MAX =

nV MAX 7.384 = => 100 nV MAX+10.207

nV = 0.814,

Solving, nV MAX = 0.814

nLIQ = 2.1 - 0.814 = 1.286

Fuel values from table A.10 and the rest from Table A.9 HR = 0.9(-74 873) + 0.1(-84 740) = -75 860 kJ/kmol fuel For the liquid water add difference (40oC – 25oC) from steam tables HLIQ = 1.286[-285 830 + 18.015(167.6 - 104.9) ] = -366 125 kJ/kmol fuel HGAS = 1.1(-393 522 + 562) + 0.814(-241 826 + 504 ) + 0.215(441) + 8.892(437) = -624 711 kJ/kmol fuel QCV = HP - HR = (-366 125 - 624 711) + 75 860 = -914 976 kJ/kmol fuel b) TP = 250 oC HP = 1.1(-393 522 + 9346) + 2.1(-241 826 + 7740) + 0.215(6808) + 8.892(6597) = -854 050 kJ/kmol QCV = HP - HR = -778 190 kJ/kmol fuel



15.43

Pentene, C5H10 is burned with stoichiometric pure oxygen in a steady flow process. After giving some heat out the products are at 700 K and used in a heat exchanger, where they are cooled to 25 °C. Find the specific heat transfer in the heat exchanger. C5H10 + νO O2 → 5 CO2 + 5 H2O 2

⇒

νO

2

= 7.5

The heat exchanger cools the products so energy equation is 5 n. F h-CO + 5 n. F h-H

2O

2

. = 5 n. h-° +Q F f CO

2

. + (5 - x) nF h°liq H

2O

. + (x) nF h°vap H

2O

Check for condensation amount Find x:

yH

2O max

=

Pg(25°) Ptot

= 0.0313 =

x 5+x

⇒

x = 0.1614

Out of the 5 H2O only 0.1614 still vapor. Find the enthalpy at 700 K in table A.9 and the formation enthalpy in table A.10 . Q = -5 ∆hCO ,700 + (5-x)(h°f liq - h°f vap - ∆h700) + x(h°f vap - h°f vap - ∆h700) n⋅ 2 F

= -5(17 761) + 4.84(-44 011 – 14 184) - 0.16(14 184) = -372 738 kJ/kmol Fu



15.44

A rigid container has a 1:1 mole ratio of propane and butane gas together with a stoichiometric ratio of air at P o, To. The charge burns and there is heat transfer to a final temperature of 1000 K. Find the final pressure and the heat transfer per kmol of fuel mixture. C3H8 + C4H10 + νO (O2 + 3.76 N2) 2

→ 7 CO2 + 9 H2O + 3.76 νO

2

N2

The carbon and hydrogen balance has been used so oxygen gives

νO2 = 7 + 9/2 = 11.5 ; νN2

= 43.24

The energy equation is

−

−

UP – UR = Q – 0 = (HP – HR) – nPRTP + nRRTR o

o

−

−

= HP + ∆HP – HR – nPRTP + nRRTR nR = 1 + 1 + 11.5 × 4.76 = 56.74,

nP = 7 + 9 + 43.24 = 59.24

From Table A.9

∆HP = 7 (33 397) + 9 (26 000) + 43.24 (21463) = 1 395 839 kJ From Table 15.3 (convert kJ/kg to kJ/kmol with M) o

o

HP – HR = 44.094 (-46 352) + 58.124 (-45 714) = - 4 700 926 kJ Substitute into energy equation Q = - 4 700 926 + 1 395 839 – 8.3145(59.24

× 1000 – 56.74 × 298.15)

= - 3 656 981 kJ/2 kmol fuel = - 1 828 490 kJ/kmol fuel



15.45

A rigid vessel initially contains 2 kmol of carbon and 2 kmol of oxygen at 25 °C, 200 kPa. Combustion occurs, and the resulting products consist of 1 kmol of carbon dioxide, 1 kmol of carbon monoxide, and excess oxygen at a temperature of 1000 K. Determine the final pressure in the vessel and the heat transfer from the vessel during the process. 2 C + 2 O2

→ 1 CO2 + 1 CO + 12 O2

Process

V = constant, C: solid, n1(GAS) = 2, n2(GAS) = 2.5 n2T2 2.5 × 1000 P2 = P1 × = 200 × = 838.4 kPa n1T1 2 × 298.2 H1 = 0 H2 = 1(-393 522 + 33 397) + 1(-110 527 + 21 686) + (1/2)(0 + 22 703) = -437 615 kJ

−

−

Q2 = (U2 - U1) = (H2 - H1) - n2RT2 + n1RT1

1

= (-437 615 - 0) - 8.3145(2.5 × 1000 - 2 × 298.2) = -453 442 kJ

COMBUSTION GAS cb



15.46

A closed, insulated container is charged with a stoichiometric ratio of oxygen and hydrogen at 25°C and 150 kPa. After combustion, liquid water at 25 °C is sprayed in such that the final temperature is 1200 K. What is the final pressure? 1

H2 + 2(O2) → H2O

Combustion reaction: Products:

1 H2O + xiH2O

° 3U2 - U1 = xihi = xihf liq = (1 + xi)HP - HR - (1 + xi)RTP + 2RTR

From Table A.9:

HR = 0, HP = -241 826 + 34 506 = -207 320 kJ/kmol °

From Table A.10: hf liq= -285 830 kJ/kmol Substitute xi(-285830 + 207320 + 8.3145 × 1200) = -207 320 - 8.3145 1200 - 2×298.15 = -213 579 3

xi = 3.116 P1V1 = nRRT1, P2V1 = npRTp P = 2

P1(1 + xi)TP

(T1)

3 2

=

150 × 4.116 × 1200 3 2

= 1657 kPa

× 298.15



15.47

In a gas turbine natural gas (methane) and stoichiometric air flows into the combustion chamber at 1000 kPa, 500 K. Secondary air, see Fig.P15.84, also at 1000 kPa, 500 K is added right after the combustion to result in a product mixture temperature of 1500 K. Find the A/F ratio mass basis for the primary reactant flow and the ratio of the secondary air to the primary air (mass flow rates ratio). CH4 + ν(O2 + 3.76 N2) →

Fuel

CO2 + 2H2O + 3.76 ν Ν2

Air1

O balance: ν = 2,

Air2

νN2 = 7.52

For the primary flow:

(A/F) mass =

Tad 1500 K

2 × 32 + 7.52 × 28.013 = 17.12 16.043

Do a C.V. around the entire set-up, then the energy equation becomes °

HR1 + HR2 = HP 1500 = HP + ∆HP 1500

Energy Eq.:

° -o HR1 = HR + ∆HR1 = hf fuel + ∆HFuel + ∆Hair1

∆Hair1 = 2(∆HO2 + 3.76 ∆HN2) = 2(6086 + 3.76 × 5911) = 56 623 kJ/kmol HR2 = 0 + ∆Hair2 = νO2 add (∆HO2 + 3.76 ∆HN2) =

νO2 add 28 311 kJ/kmol

∆HFuel ≈ M CP ∆T = 16.043 × 2.254 (500 – 298) = 7304.5 kJ/kmol ° -o hf fuel - HP = HV = 16.043 × 50 010 = 802 310 kJ/kmol

-

-

-

-

-

∆HP 1500 = ∆hCO2 + 2∆hH2O + 7.52 ∆hN2 + νO2 add (∆hO2 + 3.76 ∆hN2) = 61 705 + 2 × 48 149 + 7.52

× 38 405 + νO2 add (40 600

+ 3.76 × 38 405) = 446 809 +

νO2 add 185 003 kJ/kmol

Now substitute everything into energy equation 802 310 + 7304.5 + 56 623 +

νO2 add 28 311 = 446 809 + νO2 add 185 003

νO2 add = 2.677

mair2/mair1 = nair2/nair1 = νO2 add/ ν

= 2.677/2 =



15.48

Methane, CH4, is burned in a steady adiabatic flow process with two different Pure oxygen, O2 and case B: A mixture of O2 + x Ar. The

oxidizers: Case A:

reactants are supplied at T0, P0 and the products for both cases should be at 1800 K. Find the required equivalence ratio in case (A) and the amount of Argon, x, for a stoichiometric ratio in case (B).

a) Stoichiometric has ν = 2, actual has: CH4 + νO2 → CO2 + 2H2O + (ν - 2)O2 °

°

HR = HP + ∆HP 1800

Energy eq.: °

°

-o

-o

-o

∆HP 1800 = HR - HP = hf fuel + 0 - hf CO2 - 2hf H2O - 0 -

RHS = –74 873 –(–393 522) – 2(–241 826) = 802 301 kJ/kmol ∆hH O = 62 693, ∆hO = 51 674 all in kJ/kmol

∆hCO = 79 432, 2

2

2

∆HP 1800 = ∆hCO + 2∆hH 2

+ (ν - 2) ∆hO O

2

2

= 79 432 + 2 × 62 693 + (ν - 2) × 51 674 = 101 470 + ν 51 674 = 802 301 kJ/kmol fuel

⇒

Φ=

AFS

2

= 0.1475 AF 13.56 b) CH4 + 2 O2 + 2x Ar → CO2 + 2H2O + 2x Ar _ ∆hAr = CP Ar ∆T = CP Ar MAr ∆T from table A.5 = 13.56,

=

∆HP 1800= 79 432 + 2 × 62 693 + 2x × 0.52 × 39.948(1800 - 298) = 204 818 + x 62 402 Now the energy equation becomes 802 301 = 204 818 + x 62 402

⇒

x = 9.575



15.49

Gaseous propane mixes with air, both supplied at 500 K, 0.1 MPa. The mixture goes into a combustion chamber and products of combustion exit at 1300 K, 0.1 MPa. The products analyzed on a dry basis are 11.42% CO 2, 0.79% CO, 2.68% O2, and 85.11% N2 on a volume basis. Find the equivalence ratio and the heat transfer per kmol of fuel. C3H8 + α O2 + 3.76 α N2 → β CO2 + γ H2O + 3.76 α N2

β = 3, γ = 4, α = β + γ/2 = 5,

(A/F)S = 4.76α = 23.8

The actual combustion reaction is xC3H8 + α O2 + 3.76 α N2 → 11.42 CO2 + y H2O + 85.11 N2 + 0.79 CO + 2.68 O2 C balance:

3x = 11.42 + 0.79

H balance:

8x = 2y

=>

O balance:

2α = 2 × 11.42 + y + 0.79 + 2

N balance:

3.76 α = 85.11

=> =>

x = 4.07 y = 4x = 16.28

× 2.68 = 45.27

α = 22.6356

=>

α = 22.635

checks close enough

Rescale the equation by dividing with x to give C3H8 + 5.5614 (O2 + 3.76 N2) → 2.806 CO2 + 4 H2O + 20.91 N2 + 0.194 CO + 0.6584 O2 A/F = 5.5614 (1 + 3.76) / 1 = 26.472

φ = (A/F)S/(A/F) = 23.8 / 26.472 = o q = h P - hR = hP + o hR = hf fuel +

0.899,

%Theo. air = 1/φ = 111%

∑νi∆h(1300 K) - hR

∆hfuel + 5.5614 ∆hO2 + 20.91 ∆hN2

= -103 900 + 1.679 × 44.094 (500 – 298) + 5.5614 (6086) + 20.91 (5911) = 68 500 kJ/kmol fuel o hP = hP +

∑νi∆h(1300 K)

= 2.806 (-393 522 + 50 148) + 4(-241 826 + 38 941) + 20.91 (0 + 31 503) + 0.194 (-110 527 + 31 867) + 0.6584 (0 + 33 345) = –1 109 625 kJ/kmol fuel q = –1 109 625 – 68 500 = – 1 178 125 kJ/kmol fuel



Enthalpy of Combustion and Heating Value



15.50

Find the enthalpy of combustion and the heating value for pure carbon. C + νO [ O2 + 3.76 N2 ] → CO2 + νO 3.76 N2

Reaction:

2

oxygen balance: ° HRP

o = HP

−

νO

2

2

=1

o -° -° HR = hf CO2 – hf C = –393 522 – 0

= –393 522 kJ/kmol = –32 791 kJ/kg

(M = 12.011)

HV = 393 522 kJ/kmol = 32 791 kJ/kg



15.51

Phenol has an entry in Table 15.3, but it does not have a corresponding value of the enthalpy of formation in Table A.10. Can you calculate it? C6H5OH + νO [ O2 + 3.76 N2 ] → 3 H2O + 6 CO2 + 3.76νO N2 2

2

The C and H balance was introduced (6 C’s and 6 H’s). At the reference condition the oxygen and nitrogen have zero enthalpy of formation. o

Energy Eq.:

o

HP = HR = HP = HR

since ref. T is assumed.

o o ° -° -° -° HRP = HP − HR = HP − HR = 3 hf H2O + 6 hf CO2 – hf fuel

Table 14.3 is on a mass basis and let us chose liquid fuel, so we get the molecular weight from the composition M = 6 × 12.011 + 3 × 2.016 + 16 = 94.114 °

HRP = 94.114 (–31 117) = –2 928 545 kJ/kmol Solve the energy equation for fuel formation enthalpy ° -° -° -° hf fuel = 3 hf H2O + 6 hf CO2 – HRP

= 3 (–241 826) + 6(–393 522) – (–2 928 545) = –158 065 kJ/kmol For fuel as vapor we get °

HRP = 94.114 (–31 774) = –2 990 378 kJ/kmol ° -° -° -° hf fuel = 3 hf H2O + 6 hf CO2 – HRP

= 3 (–241 826) + 6(–393 522) – (-2 990 378) = –96 232 kJ/kmol °

Notice if I took liquid water in products to do HRP then I must use liquid value -° for hf H2O = – 285 830 kJ/kmol and the final result is the same.



15.52

Some type of wood can be characterized as C 1H1.5O0.7 with a lower heating value of 19 500 kJ/kg. Find its formation enthalpy. C1H1.5O0.7 + νO [ O2 + 3.76 N2 ] → 2

O balance: 0.7 + 2νO = 2 + 0.75 2

CO2 + 0.75 H2O + 3.76νO N2 2



νO2 = 1.025, νN2

= 3.854

M = 1 × 12.0111 + 1.5 × 1.008 + 0.7 × 16 = 24.712 °

o

o

HRP = HP − HR = –M × HV = –24.712 × 19 500 = –481 884 kJ/kmol -° -° -° = 0.75 hf H2O + 1 hf CO2 – hf fuel -° -° -° hf fuel = 0.75 hf H2O + 1 hf CO2 + M × HV = 0.75 (-241 826) + (-393 522) + 481 884 = -93 008 kJ/kmol



15.53

Do problem 15.36 using table 15.3 instead of Table A.10 for the solution. One alternative to using petroleum or natural gas as fuels is ethanol (C 2H5OH), which is commonly produced from grain by fermentation. Consider a combustion process in which liquid ethanol is burned with 120% theoretical air in a steady flow process. The reactants enter the combustion chamber at 25 °C, and the products exit at 60°C, 100 kPa. Calculate the heat transfer per kilomole of ethanol. C2H5OH + 1.2 × 3 (O2 + 3.76 N2)

→

2 CO2 + 3 H2O + 0.6 O2 + 13.54 N2

Products at 60°C, 100 kPa, so check for condensation of water yH2O MIX =

nV MAX 19.94 = 100 nV MAX+2+0.6+13.54

=> nV MAX = 4.0 > 3

⇒

No liq.

Fuel: table 15.3 select (liquid fuel, water vapor) and convert to mole basis °

HRP = 46.069 (-26 811) = -1 235 156 kJ/kmol Since the reactants enter at the reference state the energy equation becomes o

o

°

QCV = HP - HR = HP + ∆HP - HR = HRP + ∆HP ∆HP = 2 ∆hCO + 3 ∆hH O + 0.6 ∆hO + 13.54 ∆hN 2

2

2

2

= 2(1327) + 3(1178) + 0.6(1032) + 13.54(1020) = 20 618 kJ/kmol QCV = -1 235 156 + 20 618 = -1 214 538 kJ/kmol fuel



15.54

Liquid pentane is burned with dry air and the products are measured on a dry basis as: 10.1% CO2, 0.2% CO, 5.9% O2 remainder N2. Find the enthalpy of formation for the fuel and the actual equivalence ratio.

νFuC5H12 + νO

O2 + 3.76 νO N2 →

2

2

x H2O + 10.1 CO2 + 0.2 CO + 5.9 O2 + 83.8 N2 Balance of C: 5 νFu = 10.1 + 0.2 Balance of H: 12 νFu = 2 x

⇒

⇒ νFu = 2.06 x = 6 νFu = 12.36

Balance of O: 2 νO = x + 20.2 + 0.2 + 2 2

Balance of N: 2 × 3.76 νO = 83.8 × 2 2

νO

2

× 5.9 ⇒ νO

2

⇒ νO

2

= 22.28

= 22.287

OK

for 1 kmol fuel = 10.816

φ = 1,

C5H12 + 8 O2 + 8 × 3.76 N2 → 6 H2O + 5 CO2 + 30.08 N2

° ° ° -° -° -° HRP = HP - HR = 6 hf H2O + 5 hf CO2 - hf fuel °

14.3: HRP = 44 983 × 72.151

⇒

-° hf fuel = -172 998 kJ/kmol

φ = AFs / AF = νO2 stoich/νO2 AC = 8/10.816 = 0.74



15.55

Agriculturally derived butanol, C4H10O, molecular mass 74.12, also called biobutanol has a lower heating value LHV = 33 075 kJ/kg for liquid fuel. Find its formation enthalpy. The reaction equation for stiochiometric ratio is: C4H10O +

νO2 (O2 + 3.76 N2)

=> 4CO2 + 5 H2O + νO2 × 3.76 N2

where the carbon and hydrogen balance have been done. Now the oxygen O:

1 + 2νO2 = 4 × 2 + 5

=>

νO2 = 6; νN2 = 22.56

The lower heating value per kmol fuel is o o ° -° -° -° LHV = − HRP = HR − HP = hf fuel − 4 hf CO2 − 5 hf H2O

so the fuel formation enthalpy becomes -° -° -° hf fuel = LHV + 4 hf CO2 + 5 hf H2O = 74.12 × 33 075 + 4 ( −393 522 ) + 5( −241 826) = 2 451 519 – 1 574 088 – 1 209 130 = –331 699 kJ/kmol



15.56

Do Problem 15.38 using Table 15.3 instead of Table A.10 for the solution. As an alternative fuel consider liquid methanol burned with stoichiometric air both supplied at Po, To in a constant pressure process exhausting the products at 900 K. What is the heat transfer per kmol of fuel.

CH3OH + 1.5 (O2 + 3.76 N2) → 1 CO2 + 2 H2O + 5.64 N2 Reactants at 25oC, products are at 900 K, 100 kPa. CH3OH: From table 15.3 for liquid fuel, water vapor °

HRP = –19 910 kJ/kg = –19 910 × 32.042 = – 637 956 kJ/kmol fuel A9:

∆HP = 1 × 28 030 + 2 × 21 937 + 5.64 × 18 223 = 174 682 kJ/kmol fuel °

Q = HPo + ∆HP - HRo = HRP + ∆HP = – 637 956 + 174 682 = –463 274 kJ/kmol fuel



15.57

Wet biomass waste from a food-processing plant is fed to a catalytic reactor, where in a steady flow process it is converted into a low-energy fuel gas suitable for firing the processing plant boilers. The fuel gas has a composition of 50% methane, 45% carbon dioxide, and 5% hydrogen on a volumetric basis. Determine the lower heating value of this fuel gas mixture per unit volume. For 1 kmol fuel gas, 0.5 CH4 + 0.45 CO2 + 0.05 H2 + 1.025 O2

→ (0.5 + 0.45) CO2 + 1.05 H2O The lower heating value is with water vapor in the products. Since the 0.45 CO2 cancels, hRP = 0.5(-393 522) + 1.05(-241 826) - 0.5(-74 873) - 0.05(0) = -413242 kJ/kmol fuel gas With

n 100 = P/RT = = 0.04033 kmol/m3 V 8.3145 × 298.2

LHV = +413 242 × 0.04033 = 16 666 kJ/m3



15.58

Determine the lower heating value of the gas generated from coal as described in Problem 15.30. Do not include the components removed by the water scrubbers. The gas from problem 15.30 is saturated with water vapor. Lower heating value LHV has water as vapor. °

°

°

LHV = -HRP = HP - HR Only CH4, H2 and CO contributes. From 15.12 the gas mixture after the scrubbers has ∑νi = 81.9 of composition: 0.3 CH4 + 29.6 H2 + 41 CO + 10 CO2 + 0.8 N2 + 0.2016 H2O -° -° -° LHV = -[0.3HRPCH4 + 29.6HRPH2 + 41HRPCO]/81.9 = -[0.3(-50 010 × 16.043) + 29.6(-241 826) + 41(-393 522 + 110 527)]/81.9 = 232

kJ 009 kmol gas



15.59

In a picnic grill, gaseous propane and stoichiometric air are mixed and fed to the burner both at ambient Po and To. After combustion the products cool down and at some point exit at 500 K. How much heat transfer was given out for 1 kg of propane? The reaction equation for stoichiometric mixture is: C3H8 + νO2 (O2 + 3.76 N2)

→


C balance: 3 = b ; H balance: 8 = 2a O balance:

=>

a=4

2 νO2 = a + 2b = 4 + 2 × 3

νO2 = 5

=>

3.76 νO2 = c = 18.8

N2 balance:

A/F = 5 × 4.76 × 28.97 / 44.094 = 15.637 The reaction is C3H8 + 5 (O2 + 3.76 N2)

→ 4 H2O + 3 CO2 + 18.8 N2

All the water is in the vapor phase (500 K products). Energy Eq.: A9:

HR = Qout + HP

=>

o

o

HR = Qout + HP + ∆HP

∆HP = 4 × 6922 + 3 × 8305 + 18.8 × 5911 = 174 682 kJ/kmol fuel = (174 682 / 44.094) kJ/kg = 3961.6 kJ/kg

Table 15.3:

°

−HRP = HV = 46 352 kJ/kg

o Qout = HR

o

°

− HP − ∆HP = −HRP − ∆HP = HV − ∆HP = (46 352 − 3961.6) kJ/kg fuel = 42390 kJ/kg fuel



15.60

Do problem 15.40 using table 15.3 instead of Table A.10 for the solution. The reaction equation for stiochiometric ratio is:

νO2 (O2 + 3.76 N2)

C7H16 +

=> 7CO2 + 8 H2O + νO2 × 3.76 N2

So the balance (C and H was done in equation) of oxygen gives v O2 = 7 + 4 = 11, and actual one is 11×1.25 = 13.75. Now the actual reaction equation is: C H

+ 13.75 O + 51.7 N

7 16

2

=> 7CO + 8 H O + 51.7 N + 2.75 O 2

2

2

2

To find the heat transfer take a control volume as combustion chamber and heat exchanger HR + Q = HP

=>

o

o

2

°

Q = HP + ∆HP - HR = HRP + ∆HP

Now we get the enthalpy of combustion from table 14.3, which is per kg, so scale it with the molecular weight for the fuel. Add all the ∆HP from A.9 ∆HP = 7 ∆hCO + 8 ∆hH O + 51.7 ∆hN + 2.75 ∆hO 2

2

2

2

= 7(12 906) + 8(10 499) + 51.7(8894) + 2.75(9245) = 659 578 °

Q = M HRP +

kJ kmol

∆HP = 100.205(-44 922) + 659 578

= - 3 841 831 kJ/kmol fuel



15.61

Propylbenzene, C9H12, is listed in Table 15.3, but not in table A.10. No molecular weight is listed in the book. Find the molecular weight, the enthalpy of formation for the liquid fuel and the enthalpy of evaporation. C9H12 + 12 O2 → 9 CO2 + 6 H2O ^ = 9 × 12.011 + 6 × 2.016 = 120.195 M o o -° h ==H -H = RP

P

-°

ν i hf

∑P

R

i

-° - hf

⇒

Fu

-° hf Fu

-° -° -h

νihf

=

∑P

i

RP

Formation enthalpies from Table A.10 and enthalpy of combustion from Table 15.3 -° -° -° ^ (-41 219) hfFu = 9hfCO2 + 6hfH2O g - M liq

Fu H2O vap

= 9(-393 522) + 6(-241 826) – 120.195(-41 219) = -38 336 kJ/kmol Take the enthalpy of combustion from Table 15.3 for fuel as a gas and as a vapor, the difference is the enthalpy of evaporation °

°

hfg = -(hRP gas - hRP liq ) = 41 603 – 41 219 = 384 kJ/kg



15.62

Consider natural gas A which are listed in Table 15.2. Calculate the enthalpy of combustion of each gas at 25 °C, assuming that the products include vapor water. Repeat the answer for liquid water in the products. Natural Gas A 0.939 CH4 + 0.036 C2H6 + 0.012 C3H8 + 0.013 C4H10 + 2.1485 O2 + 3.76 × 2.1485 N2 → 1.099 CO2 + 2.099 H2O + 8.0784 N2 HR = 0.939(-74 878) + 0.036(-84 740) + 0.012(-103 900) + 0.013(-126 200) = -76244 kJ a) vapor H2O HP = 1.099(-393 522) + 2.099(-241 826) = -940 074 kJ/kmol hRP = HP - HR = -863 830 kJ/kmol b) Liq. H2O HP = 1.099(-393 522) + 2.099(-285 830) = -1 032 438 hRP = -956 194 kJ/kmol



15.63

Consider natural gas D which are listed in Table 15.2. Calculate the enthalpy of combustion of each gas at 25 °C, assuming that the products include vapor water. Repeat the answer for liquid water in the products. Natural Gas D: 0.543 CH4 + 0.163 C2H6 + 0.162 C3H8 + 0.074 C4H10 + 0.058 N2 + O2 + N2

→

1.651 CO2 + 2.593 H2O + N2

HR = 0.543(-74 873) + 0.163(-84 740) + 0.162(-130 900) + 0.074(-126 200) = -80 639 kJ a) vapor H2O HP = 1.651(-393 522) + 2.593(-241 826) = -1 276 760 kJ hRP = -1 196 121 kJ/kmol b) Liq. H2O HP = 1.651(-393 522) + 2.593(-285 830) = -1 390 862 kJ hRP = -1 310 223 kJ/kmol



15.64

Gaseous propane and stoichiometric air are mixed and fed to a burner both at ambient Po and To. After combustion the products cool down and eventually reach ambient To. How much heat transfer was given out for 1 kg of propane? The reaction equation for stoichiometric mixture is: C3H8 + νO2 (O2 + 3.76 N2)

→


C balance: 3 = b ; H balance: 8 = 2a O balance:

a=4

2 νO2 = a + 2b = 4 + 2 × 3 =>

N2 balance:

=>

νO2 = 5

3.76 νO2 = c = 18.8

A/F = 5 × 4.76 × 28.97 / 44.094 = 15.637 The reaction is C3H8 + 5 (O2 + 3.76 N2)

→ 4 H2O + 3 CO2 + 18.8 N2

First we solve the problem assuming all the water is in the vapor phase and then we can solve with the proper split of watero into liquid and o vapor masses. Energy Eq.: HR = Qout + HP => HR = Qout + HP o

o

°

Qout = HR − HP = −HRP = HV = 46 352 kJ/kg fuel At To we need to check for condensation of water partial P limited to P g. nv max 3.169 = 100 nv max + 3 + 18.8 nv max = 0.7135 and then nliq = 4 - nv max = 3.2865 yv max =

⇒

Pg P

=

So now we know the split of the 4 moles of water into vapor and liquid. The change in the heating value becomes ( HV liq water = 50 343 kJ/kg fuel) Qout = HV =

0.7135 3.2865 46 352 + 50 343 = 49 631 kJ/kg fuel 4 4



15.65

Blast furnace gas in a steel mill is available at 250 °C to be burned for the generation of steam. The composition of this gas is, on a volumetric basis, Component CH4 H2 CO CO2 N2 H 2O Percent by volume 0.1 2.4 23.3 14.4 56.4 3.4 Find the lower heating value (kJ/m 3) of this gas at 250°C and ambient pressure. Of the six components in the gas mixture, only the first 3 contribute to the heating value. These are, per kmol of mixture: 0.024 H2, 0.001 CH4, 0.233 CO For these components, 0.024 H2 + 0.001 CH4 + 0.233 CO + 0.1305 O2

→ 0.026 H2O + 0.234 CO2 The remainder need not be included in the calculation, as the contributions to reactants and products cancel. For the lower HV(water as vapor) at 250 °C hRP = 0.026(-241 826 + 7742) + 0.234(-393 522 + 9348) - 0.024(0 + 6558) - 0.001(-74 873 + 2.254 × 16.04(250-25)) - 0.233(-110 527 + 6625) – 0.1305(0 + 6810) kJ

= -72 573 kmol fuel

− v- = R To/Po = 8.3145 × 523.2/100 = 43.5015 m3/kmol 0

LHV = +72 573 / 43.5015 = 1668 kJ/m3



15.66

A burner receives a mixture of two fuels with mass fraction 40% n-butane and 60% methanol, both vapor. The fuel is burned with stoichiometric air. Find the product composition and the lower heating value of this fuel mixture (kJ/kg fuel mix). Since the fuel mixture is specified on a mass basis we need to find the mole fractions for the combustion equation. From Eq.13.4 we get ybutane = (0.4/58.124) / [0.4/58.124 + 0.6/32.042] = 0.26875 ymethanol = 1 – ybutane = 0.73125 The reaction equation is 0.73125 CH3OH + 0.26875 C4H10 + νO2 (O2 + 3.76 N2)

→ νCO2 CO2 + νH2OH2O +

3.76 νO2 N2

C balance:

0.73125 + 4 × 0.26875 = νCO2 = 1.80625

H2 balance:

2 × 0.73125 + 5 × 0.26875 = νH2O = 2.80625

O balance:

0.73125 + 2 νO2 = 2 νCO2 + νH2O = 6.41875 => νO2 = 2.84375

Now the products are: 1.80625 CO2 + 2.80625 H2O + 10.6925 N2 Since the enthalpy of combustion is on a mass basis in table 15.3 (this is also the negative of the heating value) we get LHV = 0.4 × 45 714 + 0.6 × 21 093 = 30 941 kJ/kg fuel mixture Notice we took fuel vapor and water as vapor (lower heating value).



15.67

In an experiment a 1:1 mole ratio propane and butane is burned in a steady flow with stoichiometric air. Both fuels and air are supplied as gases at 298 K and 100 kPa. The products are cooled to 1000 K as they give heat to some application. Find the lower heating value (per kg fuel mixture) and the total heat transfer for 1 kmol of fuel mixture used. Combustion: C3H8 + C4H10 + 11.5(O2 + 3.76 N) → 7 CO2 + 9 H2O + 43.24 N2

νO2 = 7 + 9/2 = 11.5

=>

νN2 = 3.76 × 11.5 = 43.24

The enthalpy of combustion for the 2 kmol of fuel becomes from Tbl 15.3 o

o

HP – HR = (-46 352) 44.094 + (-45 714) 58.124 = - 4 700 925.6 kJ/2 kmol fuel °

HV = –HRP =

4 700 925.6 = 45 989 kJ/kg fuel Convert to mass 44.094 + 58.124

basis

∆HP = 7 × 33 397 + 9 × 26 000 + 43.24 × 21 463 = 1 395 839 kJ/2 kmol fuel The heat transfer for the two kmol of fuel is °

Q = HR – HP = –HRP – ∆HP = 4 700 925.6 – 1 395 839 = 3 305 087 kJ/2 kmol fuel = 1 652 544 kJ/kmol fuel



15.68

Liquid nitromethane is added to the air in a carburetor to make a stoichiometric mixture where both fuel and air are added at 298 K, 100 kPa. After combustion a constant pressure heat exchanger brings the products to 600 K before being exhausted. Assume the nitrogen in the fuel becomes N 2 gas. Find the total heat transfer per kmole fuel in the whole process. CH3NO2 +

νO2 (O2 + 3.76 N2) →

1.5 H2O + 1 CO2 + a N2

C and H balances done in equation. The remaining O balance: 2 + 2 νO2 = 1.5 + 2 N balance: 1 + 3.76 νO2 × 2 = 2a Energy eq.:

HR + Q = HP

=>

νO2 = 0.75

=> =>

a = 3.32 °

°

Q = HP − HR = HP − HR + ∆HP − ∆HR

The reactants enter at the reference state, ∆HR = 0, and the products at 600 K from table A.9

∆HP =

1.5 ∆hH

2O

+ ∆hCO + 3.32 ∆hN 2

2

= 1.5 (10 499) + 1 (12 906) + 3.32 (8894) = 58 183 kJ/kmol fuel °

°

°

HP − HR = HRP = 61.04 (-10 537) = -643 178 kJ/kmol Q = -643 178 + 58 183 = -584 995 kJ/kmol fuel



15.69

Natural gas, we assume methane, is burned with 200% theoretical air and the reactants are supplied as gases at the reference temperature and pressure. The products are flowing through a heat exchanger where they give off energy to some water flowing in at 20oC, 500 kPa and out at 700 oC, 500 kPa. The products exit at 400 K to the chimney. How much energy per kmole fuel can the products deliver and how many kg water per kg fuel can they heat?

The reaction equation for stoichiometric mixture is: CH4 + νO2 (O2 + 3.76 N2) → 2 H2O + 1 CO2 + c N2 2 νO2 = 2 + 2

O balance:


νO2 = 2

=>

νO2 = 2 × 2 = 4

CH4 + 4 (O2 + 3.76 N2)


→ 2 H2O + 1 CO2 + 15.04 N2 + 2 O2

The products are cooled to 400 K (so we do not consider condensation) and the energy equation is °

°

HR + Q = HP = HP + ∆HP = HR + Q

Energy Eq.:

Q= From Table 15.3:

° HP

-

° HRP

° HR

+ ∆HP =

° HRP

+ ∆HP

= 16.04 (-50 010) = -802 160 kJ/kmol

∆HP = ∆h-*CO2 + 2 ∆h-*H2O + 2 ∆h-*O2 + 15.04 ∆h-*N2 From Table A.9

∆HP 400

= 4003 + 2 × 3450 + 2 × 3027 + 15.04 × 2971 = 61 641 kJ/kmol °

Q = HRP + ∆HP = -802 160 + 61 641 = -740 519 kJ/kmol qprod = -Q / M = 740 519 / 16.04 = 46 167 kJ/kg fuel The water flow has a required heat transfer, using B.1.3 and B.1.4 as qH2O = hout – hin = 3925.97 – 83.81 = 3842.2 kJ/kg water The mass of water becomes mH2O / mfuel = qprod / qH2O = 12.0 kg water / kg fuel



15.70

An isobaric combustion process receives gaseous benzene C6H6 and air in a stoichiometric ratio at Po, To. To limit the product temperature to 2000 K, liquid water is sprayed in after the combustion. Find the kmol of liquid water added per kmol of fuel and the dew point of the combined products. The reaction equation for stoichiometric mixture with C and H balance done is: C6H6 + νO2 (O2 + 3.76 N2) O balance:

2 νO = 3 + 6 × 2 = 15

N balance:

c = 3.76

→

⇒ νO

2

νO

2

3 H2O + 6 CO2 + c N2 2

= 7.5

= 3.76 × 7.5 = 28.2

With x kmol of water added per kmol fuel the products are Products: (3 + x) H2O + 6 CO2 + 28.2 N2 Energy Eq.:

° ° -° HR = HR + x hf H2O liq = HP + ∆HP ° -° = HP + x hf H2O vap + (3 + x) ∆hH

2O

+ 6 ∆hCO + 28.2 ∆hN 2

2

Where the extra water is shown explicitly. Rearrange to get ° ° HR − HP − 6∆hCO − 28.2∆hN − 3∆hH O 2

2

2

-° -° = x (hf H2O vap − hf H2O liq + ∆hH

O 2

)

40 576 × 78.114 − 6 × 91 439 − 28.2 × 56 137 − 3 × 72 788 = x [-241 826 –(–285 830) + 72 788 ] 819 493 = x (116 792) ⇒

x = 7.017 kmol/kmol fuel

3+x = 0.2265 6 + 28.2 + 3 + x ⇒ Pv = yvP = 0.2265 × 101.325 = 22.95 kPa

Dew point: yv =

B.1.2: Tdew = 63oC



15.71

Gasoline, C7H17, is burned in a steady state burner with stoichiometric air at P o, To. The gasoline is flowing as a liquid at T o to a carburetor where it is mixed with air to produce a fuel air gas mixture at To. The carburetor takes some heat transfer from the hot products to do the heating. After the combustion the products go through a heat exchanger, which they leave at 600 K. The gasoline consumption is 10 kg per hour. How much power is given out in the heat exchanger and how much power does the carburetor need? Stoichiometric combustion: C7H17 + νO

2

(O2 + 3.76 N2) →

O balance:

2 νO = 8.5 + 14 = 22.5

N balance:

c = 3.76

2

νO

2

8.5 H2O + 7 CO2 + c N2

⇒ νO

2

= 11.25

= 3.76 × 11.25 = 42.3

MFUEL = 7 MC + 17 MH = 7 × 12.011 + 8.5 × 2.016 = 101.213 C.V. Total, heat exchanger and carburetor included, Q out. Energy Eq.:

°

°

HR = HR = HP + ∆HP + Qout

From Table A.9

∆HP = 8.5 × 10 499 + 7 × 12 906 + 42.3 From energy equation and Table 15.3 °

°

× 8894 = 555 800 kJ/kmol

°

Qout = HR - HP - ∆HP = -HRP - ∆HP = 101.213 (44 506) – 555 800 = 3 948 786 kJ/kmol Now the power output is . 10 . . Q = n Qout = Qout m/M = 3 948 786 × / 101.213 = 108.4 kW 3600 The carburetor air comes in and leaves at the same T so no change in energy, all we need is to evaporate the fuel, h fg so . 10 1 . Q = m hfg = (44 886 – 44 506) = × 380 = 1.06 kW 3600 360

Here we used Table 15.3 for fuel liquid and fuel vapor to get h fg and any phase of the water as long as it is the same for the two.



Adiabatic Flame Temperature



15.72

In a rocket, hydrogen is burned with air, both reactants supplied as gases at P o, To. The combustion is adiabatic and the mixture is stoichiometeric (100% theoretical air). Find the products dew point and the adiabatic flame temperature (~2500 K). The reaction equation is: H2 + νO2 (O2 + 3.76 N2) => H2O + 3.76 vO2 N2 The balance of hydrogen is done, now for oxygen we need vO2 = 0.5 and thus we have 1.88 for nitrogen. yv = 1/(1+1.88) = 0.3472 => Pv = 101.325 × 0.3472 = 35.18 kPa = P g Table B.1.2:

Tdew = 72.6 C.

HR = HP => 0 = -241826 +

∆hwater + 1.88 ∆hnitrogen

Find now from table A.9 the two enthalpy terms At 2400 K : At 2600 K :

∆HP ∆HP

= 93741 + 1.88 × 70640 = 226544 kJ/kmol fuel = 104520 + 1.88

× 77963 = 251090 kJ/kmol fuel

Then interpolate to hit 241 826 to give

T = 2525 K



15.73

Hydrogen gas is burned with pure oxygen in a steady flow burner where both reactants are supplied in a stoichiometric ratio at the reference pressure and temperature. What is the adiabatic flame temperature? The reaction equation is: H2 + νO2 O2

=> H2O

The balance of hydrogen is done, now for oxygen we need vO2 = 0.5. Energy Eq.: =>

HR = HP => 0 = -241 826 + ∆hH2O = 241 826 kJ/kmol

∆h-H2O

Interpolate now in table A.9 for the temperature to give this enthalpy T = 4991 K



15.74

Some type of wood can be characterized as C 1H1.5O0.7 with a lower heating value of 19 500 kJ/kg. Find its adiabatic flame temperature when burned with stoichiometric air at 100 kPa, 298 K. C1H1.5O0.7 + νO [ O2 + 3.76 N2 ] → 2

O balance: 0.7 + 2νO = 2 + 0.75 2

CO2 + 0.75 H2O + 3.76νO N2 2



νO

2

= 1.025,

νN2

= 3.854

M = 1 × 12.0111 + 1.5 × 1.008 + 0.7 × 16 = 24.712 °

o

o

HRP = HP − HR = –M × HV = –24.712 × 19 500 = –481 884 kJ/kmol Energy Eq.:

°

°

HP = HP + ∆HP = HR = HR -

-

⇒ ∆HP =

°

°

°

HR – HP = –HRP

-

∆HP = ∆hCO2 + 0.75 ∆hH2O + 3.854 ∆hN2 = 481 884 kJ/kmol at 2400 K: at 2600 K:

∆HP = 115 779 + 0.75 × 93 741 + 3.854 × 70 640 = 458 331 kJ ∆HP = 128 074 + 0.75 × 104 520 + 3.854 × 77 963 = 506 933 kJ T = 2400 + 200

481 884 - 458 331 = 2497 K 506 933 - 458 331

Comment: Most wood has some water and some non-combustible solids material so the actual flame temperature will be much lower.



15.75

Carbon is burned with air in a furnace with 150% theoretical air and both reactants are supplied at the reference pressure and temperature. What is the adiabatic flame temperature? C + νO O2 + 3.76 νO N2 2

→

2

1 CO2 + 3.76 νO N2 2

From this we find νO = 1 and the actual combustion reaction is 2

→

C + 1.5 O + 5.64 N 2 °

2 °

HP = HP + ∆HP = HR = HR °

1 CO + 5.64 N + 0.5 O

2

2

2

⇒

°

∆HP = HR - HP = 0 - (-393 522) = 393 522 kJ/kmol ∆HP = ∆hCO2 + 5.64 ∆hN2 + 0.5 ∆hO2 Find T so ∆HP takes on the required value. To start guessing assume all products are nitrogen (1 + 5.64 + 0.5 = 7.14) that gives 1900 < T < 2000 K from Table A.9. ∆HP 1900 = 85 420 + 5.64 ×52 549 + 0.5×55 414 = 409 503 too high

∆HP 1800

= 79 432 + 5.64 ×48 979 + 0.5×51 674 = 381 511

Linear interpolation to find T = 1800 + 100

393 522 - 381 511 = 1843 K 409 503 - 381 511



15.76

Hydrogen gas is burned with 200% theoretical air in a steady flow burner where both reactants are supplied at the reference pressure and temperature. What is the adiabatic flame temperature? The stoichiometric reaction equation is: H2 + νO2 (O2 + 3.76 N2) => H2O + 3.76 νO2 N2 The balance of hydrogen is done, now for oxygen we need νO2 = 0.5 and thus we have for the actual mixture νO2 = 1. The actual reaction is H2 + 1 (O2 + 3.76 N2) => 1 H2O + 3.76 N2 + 0.5 O2 The energy equation with formation enthalpy from A.9 or A.10 for water is HR = HP => 0 = -241 826 +

∆hH2O + 3.76 ∆hN2 + 0.5 ∆hO2

Find now from table A.9 the two enthalpy terms At 2000 K : At 1800 K : At 1600 K : At 1700 K :

∆HP ∆HP ∆HP ∆HP

× 56 137 + 0.5 × 59 176 = 313 451 × 48 979 + 0.5 × 51 674 = 272 691 = 52 907 + 3.76 × 41 904 + 0.5 × 44 267 = 232 600 = 57 757 + 3.76 × 45 430 + 0.5 × 47 959 = 252 553 = 72 788 + 3.76 = 62 693 + 3.76

Then interpolate to hit 241 826 to give 241 826 - 232 600 T = 1600 + 100 252 553 - 232 600 = 1646 K



15.77

What is the adiabatic flame temperature before the secondary air is added in Problem 15.47 CH4 + ν(O2 + 3.76 N2) →

Fuel

CO2 + 2H2O + 3.76 ν Ν2

Air1

O balance: ν = 2,

Air2

νN2 = 7.52

Tad 1500 K

Do a C.V. around the combustion chamber, then the energy equation becomes °

HR1 = HP 1 = HP + ∆HP 1

Energy Eq.:

° -o HR1 = HR + ∆HR1 = hf fuel + ∆HFuel + ∆Hair1

∆Hair1 = 2(∆HO2 + 3.76 ∆HN2) = 2(6086 + 3.76 × 5911) = 56 623 kJ/kmol ∆HFuel ≈ M CP ∆T = 16.043 × 2.254 (500 – 298) = 7304.5 kJ/kmol ° -o hf fuel - HP = HV = 16.043 × 50 010 = 802 310 kJ/kmol

-

-

-

∆HP 1 = ∆hCO2 + 0.75 ∆hH2O + 3.854 ∆hN2 ° -o = hf fuel + ∆HFuel + ∆Hair1 - HP

(from energy Eq.)

= 802 310 + 7304.5 + 56 623 = 866 237 kJ/kmol fuel at 2400 K: at 2600 K:

∆HP 1 = 115 779 + 2 × 93 741 + 7.52 × 70 640 = 834 474 kJ ∆HP 1 = 128 074 + 2 × 104 520 + 7.52 × 77 963 = 923 396 kJ T = 2400 + 200

866 237 - 834 474 = 2471 K 923 396 - 834 474



15.78

Butane gas at 25 °C is mixed with 150% theoretical air at 600 K and is burned in an adiabatic steady flow combustor. What is the temperature of the products exiting the combustor?

C4H10 + 1.5×6.5 (O2 + 3.76 N2 )→ 4 CO2 + 5 H2O + 3.25 O2 + 36.66 N2 °

Energy Eq.: Reactants:

° -o HR = hC4H10 = hf IG = -126 200 kJ ° HP

°

HP - HR = 0 => ∆HP = HR + ∆HR - HP ∆HR = 9.75(9245) + 36.66(8894) = 416 193 kJ ; =>

HR = +289 993 kJ

= 4(-393522) + 5(-241826) = -2 783 218 kJ/kmol -*

-*

-*

-*

∆HP = 4 ∆hCO2 + 5 ∆hH2O + 3.25 ∆hO2 + 36.66 ∆hN2 From the energy equation we then get

∆HP = -126 200 + 416 193 –(-2 783 218) = 3 073 211 kJ/kmol Trial and Error:

LHS2000 K = 2 980 000,

Linear interpolation to match RHS

LHS2200 K = 3 369 866

=> TP = 2048 K



15.79

A gas turbine burns methane with 200% theoretical air. The air and fuel comes in through two separate compressors bringing them from 100 kPa, 298 K to 1400 kPa and after mixing enters the combustion chamber at 600 K. Find the adiabatic flame temperature using constant specific heat for the ∆HP terms. The reaction equation for stoichiometric mixture is: CH4 + νO2 (O2 + 3.76 N2) 2 νO2 = 2 + 2

O balance:


→ 2 H2O + 1 CO2 + c N2

νO2 = 2

=>

νO2 = 2 × 2 = 4


→ 2 H2O + 1 CO2 + 15.04 N2 + 2 O2

CH4 + 4 (O2 + 3.76 N2)

The energy equation around the combustion chamber becomes Energy Eq.:

HP - HR = 0

=>

°

°

°

∆HP = HR + ∆HR - HP = -HRP + ∆HR -

-

∆HR = ∆HFuel + ∆Hair = M CP ∆T + 4(∆hO2 + 3.76 ∆hN2) = 16.043 × 2.254 (600–298) + 4(9245 + 3.76 ° -HRP

= 16.043 × 50 010 = 802 310 kJ/kmol = ∆h

∆H P

× 8894) = 181 666 kJ/kmol

+ 2 ∆h

+ 15.04 ∆h

+ 2 ∆h

≈ ∆Τ

-

νC

∑ Eq.) = 802 310 + 181 666 = 983 976 kJ/kmol (from energy CO2

H2O

N2

O2

i Pi

∑νiC-Pi = 0.842 × 44.01 + 2 × 1.872 × 18.015 + 15.04 × 1.042 × 28.013 + 2 × 0.922 × 31.999 = 602.52 kJ/kmol-K -

∆Τ = ∆HP / ∑νiCPi = 983 976 / 602.52 = 1633.1 K T = 298 + 1633 = 1931 K



15.80

Extend the solution to the previous problem by using Table A.9 for the ∆HP terms. The reaction equation for stoichiometric mixture is: CH4 + νO2 (O2 + 3.76 N2) O balance:

2 νO2 = 2 + 2


νO2 = 2

=>

νO2 = 2 × 2 = 4

CH4 + 4 (O2 + 3.76 N2)

→ 2 H2O + 1 CO2 + c N2 so now more O2 and N2

→ 2 H2O + 1 CO2 + 15.04 N2 + 2 O2

The energy equation around the combustion chamber becomes Energy Eq.:

HP - HR = 0

=>

°

°

°

∆HP = HR + ∆HR - HP = -HRP + ∆HR -

-

∆HR = ∆HFuel + ∆Hair = M CP ∆T + 4(∆hO2 + 3.76 ∆hN2) = 16.043 × 2.254 (600–298) + 4(9245 + 3.76

× 8894) = 181 666 kJ/kmol

° -HRP

= 16.043 × 50 010 = 802 310 kJ/kmol ∆HP = ∆hCO2 + 2 ∆hH2O + 15.04 ∆hN2 + 2 ∆hO2 = 802 310 + 181 666 = 983 976 kJ/kmol (from energy Eq.) Trial and error with ∆h from Table A.9 At 1800 K ∆HP = 79 432 + 2 ×62 693 + 15.04 ×48 979 + 2 ×51 674 = 1 044 810 kJ/kmol At 1700 K ∆HP = 73 480 + 2 ×57 757 + 15.04 ×45 430 + 2 ×47 959 = 968 179 kJ/kmol Linear interpolation:

T = 1700 + 100

983 976 - 968 179 = 1721 K 1 044 810 - 968 179



15.81

A stoichiometric mixture of benzene, C6H6, and air is mixed from the reactants flowing at 25°C, 100 kPa. Find the adiabatic flame temperature. What is the error if constant specific heat at T0 for the products from Table A.5 are used? C6H6 + νO O2 + 3.76 νO N2 2

2

→

6CO2 + 3H2O + 3.76 νO N2 2

νO2 = 6 + 3/2 = 7.5 ⇒ νN2 = 28.2 °

°

H = HP + ∆H = H = HR

∆HP =

P ° -HRP

P

⇒

R

= 40576 × 78.114 = 3 169 554 kJ/kmol ∆HP = 6 ∆hCO2 + 3 ∆hH2O + 28.2 ∆hN2

∆HP 2600K = 6(128074) + 3(104 520) + 28.2(77 963) = 3 280 600, ∆HP 2400K = 6(115 779) + 3(93 741) + 28.2(70 640) = 2 968 000 Linear interpolation

⇒

TAD= 2529 K

∑νiC-Pi = 6 × 0.842 × 44.01 + 3 × 1.872 × 18.015 + 28.2 × 1.042 × 28.013 = 1146.66 kJ/kmol K -

∆T = ∆HP/∑νiCPi = 3 169 554 / 1146.66 = 2764 ⇒

TAD= 3062 K, 21% high



15.82

A gas-turbine burns natural gas (assume methane) where the air is supplied to the combustor at 1000 kPa, 500 K and the fuel is at 298 K, 1000 kPa. What is the equivalence ratio and the percent theoretical air if the adiabatic flame temperature should be limited to 1800 K? The reaction equation for a mixture with excess air is: CH4 + νO2 (O2 + 3.76 N2) HP =

° HP

→ 2 H2O + 1 CO2 + 3.76νO2 N2 + (νO2 – 2)O2 °

+ ∆HP = HR = HR + ∆HR

From table A.9 at 500 K (notice fuel is at 298 K)

∆HR = 0 + νO2(∆hO2 + 3.76 ∆hN2) = νO2(6086 + 3.76 × 5911) = 28 311.4 νO2 From table A.9 at 1800 K:

∆HP = 2 ∆hH2O + ∆hCO2 + 3.76 νO2 ∆hN2 + (νO2 – 2) ∆hO2 = 2 × 62 693 + 79432 + 3.76 = 101 470 + 235 835 From table 15.3:

° HP

-

° HR

=

° HRP

νO2 × 48 979 + (νO2 – 2) 51 674

νO2 = 16.04(-50 010) = -802 160 kJ/kmol

Now substitute all terms into the energy equation -802 160 + 101 470 + 235 835

νO2 = 28 311.4 νO2

Solve for νO2 802 160 - 101 470

νO2 = 235 835 - 28 311.4 = 3.376 %Theoretical air = 100 (3.376 / 2) = 168.8 %

Φ = AFs / AF = 2 / 3.376 = 0.592



15.83

Acetylene gas at 25°C, 100 kPa is fed to the head of a cutting torch. Calculate the adiabatic flame temperature if the acetylene is burned with a. 100% theoretical air at 25°C. b. 100% theoretical oxygen at 25 °C. C2H2 + 2.5 O2 + 2.5 × 3.76 N2 → 2 CO2 + 1 H2O + 9.4 N2

a)

-o H = hf C2H2 = +226 731 kJ/kmol

from table A.10

R

-* -* -* HP = 2(-393 522 + ∆hCO2) + 1(-241 826 + ∆hH2O) + 9.4 ∆hN2 QCV = HP - HR = 0

-* -* -* 2 ∆hCO2 + 1 ∆hH2O + 9.4 ∆hN2 = 1 255 601 kJ

⇒

Trial and Error A.9:

LHS2800 = 1 198 369,

Linear interpolation:

TPROD = 2909 K

LHS3000 = 1 303 775

C2H2 + 2.5 O2 → 2 CO2 + H2O

b)

HR = +226 731 kJ ;

⇒

-* -* HP = 2(-393 522 + ∆hCO2) + 1(-241 826 + ∆hH2O)

-* -* 2 ∆hCO2 + 1 ∆hH2O = 1 255 601 kJ/kmol fuel

At 6000 K (limit of A.9)

2 × 343 782 + 302 295 = 989 859

At 5600 K

2 × 317 870 + 278 161 = 913 901

Slope 75 958/400 K change Extrapolate to cover the difference above 989 859 kJ/kmol fuel TPROD ≈ 6000 + 400(265 742/75 958) ≈ 7400 K



15.84

Liquid n-butane at T0, is sprayed into a gas turbine with primary air flowing at 1.0 MPa, 400 K in a stoichiometric ratio. After complete combustion, the products are at the adiabatic flame temperature, which is too high, so secondary air at 1.0 MPa, 400 K is added, with the resulting mixture being at 1400 K. Show that Tad > 1400 K and find the ratio of secondary to primary air flow. C.V. Combustion Chamber. C4H10 + 6.5 O2 + 6.5 × 3.76 N2 → 5 H2O + 4 CO2 + 24.44 N2 Primary Air

Fuel

Secondary

To turbine

Combustion Chamber Energy Eq.: ° HP

Air

Mixing

TAD

1400 K

Hair + Hfuel = HR = HP °

+ ∆HP = HR + ∆HR

⇒

°

°

°

∆HP = HR + ∆HR - HP = -HRP + ∆HR

∆HP = 45344 × 58.124 + 6.5(3.76 × 2971 + 3027) = 2 727 861 kJ/kmol ∆HP 1400 Remark:

= 5 × 43491 + 4 × 55895 + 24.44 × 34936 = 1 294 871 < HP

Try TAD > 1400:

∆HP = 2 658 263 @2400 K, ∆HP = 2 940 312 @2600 K C.V. Mixing Chamber. Air Second: νO2 sO2 + 3.76 N2 ∆ΗP + νO2 second ∆Hair = ∆HP 1400 + νO2 second ∆Hair 1400

∆HP - ∆HP 1400 ⇒ νO2 second= ∆H - ∆H air 1400

air 400

=

1432990 = 9.3 168317 - 14198

ratio = νO2 sec/νO2 prim = 9.3/6.5 = 1.43



15.85

Ethene, C2H4, burns with 150% theoretical air in a steady flow constant-pressure process with reactants entering at P0, T0. Find the adiabatic flame temperature. Stoichiometric C2H4 + 3(O2 + 3.76N2) → 2CO2 + 2H2O + 11.28N2 Actual C2H4 + 4.5(O2 + 3.76N2) → 2CO2 + 2H2O + 1.5 O2 + 16.92N2 ° HP = HP + 2∆h-CO + 2∆h-H 2

° HR

2O

-° = hf Fu

+ 1.5∆h-O + 16.92∆h-N 2

∆HP +

° HP

=

2

° HR

°

kJ

⇒ ∆HP = -HRP = 28.054 × 47158 = 1 322 970.5 kmol Fu -

∆HP = 2∆hCO

2

+ 2∆hH

2O

+ 1.5∆hO + 16.92∆hN 2

Initial guess based on (2+2+1.5+16.92) N2 from A.9:

∆HP(2000) = 1 366 982, => TAD

2

T1 = 2100 K

∆HP(1900) = 1 278 398

1950 K



15.86

Natural gas, we assume methane, is burned with 200% theoretical air and the reactants are supplied as gases at the reference temperature and pressure. The products are flowing through a heat exchanger and then out the exhaust, as in Fig. P15.86. What is the adiabatic flame temperature right after combustion before the heat exchanger? The reaction equation for stoichiometric mixture is: CH4 + νO2 (O2 + 3.76 N2) 2 νO2 = 2 + 2

O balance:


° HP

1 CO2 + 2 H2O + c N2

νO2 = 2

=>

νO2 = 2 × 2 = 4

CH4 + 4 (O2 + 3.76 N2) Energy Eq.:

→

→


1 CO2 +

2 H2O + 15.04 N2 + 2 O2

Hair + Hfuel = HR = HP °

+ ∆HP = HR + ∆HR

From Table 14.3:

⇒

°

°

°

∆HP = HR + ∆HR - HP = -HRP + 0

°

-HRP = -16.04 (-50 010) = 802 160 kJ/kmol -*

-*

-*

-*

∆HP = ∆hCO2 + 2 ∆hH2O + 2 ∆hO2 + 15.04 ∆hN2 From Table A.9

∆HP 1600 ∆HP 1500

= 67 659 + 2 × 52 907 + 2 × 44 267 + 15.04 = 61 705 + 2 × 48 149 + 2 × 40 600 + 15.04

× 41 904 = 892 243 × 38 405 = 816 814

∆HP 1400

= 55 895 + 2 × 43 491 + 2 × 36 958 + 15.04

× 34 936 = 742 230

Linear interpolation to get 802 160 T = 1400 + 100

802 160 - 742 230 = 1480 K 816 814 - 742 230



15.87

Solid carbon is burned with stoichiometric air in a steady flow process. The reactants at T0, P0 are heated in a preheater to T2 = 500 K as shown in Fig. P15.87, with the energy given by the product gases before flowing to a second heat exchanger, which they leave at T0. Find the temperature of the products T4, and the heat transfer per kmol of fuel (4 to 5) in the second heat exchanger. Combustion of carbon: C + O2 + 3.76 N2 → CO2 + 3.76 N2 C.V. Combustion chamber and preheater from 1 to 4, no external Q. For this CV states 2 and 3 are internal and do not appear in equations. Energy Eq.: ° ° HR = HR = HP = HP + ∆HP = hf CO + ∆hCO + 3.76∆hN 4

Table A.9 or A.10:

∆HP4 ∆HP

=

4

2

2

2

° hf CO = -393 522 kJ/kmol, reference HR = 0 2

° HR

4 2400

°

-

− HP = 0 − hf CO2 = 393 522 kJ/kmol fuel

= 115 779 + 3.76 × 70 640 = 381 385 kJ/kmol fuel,

∆HP4 2600 = 128 074 + 3.76 × 77 963 = 421 215 kJ/kmol fuel ⇒

interpolate

T4 = T

= 2461 K ad.flame

Control volume: Total. Then energy equation: °

−

°

HR + Q = HP kJ -° − -° Q = HRP = hf CO2 - 0 = -393 522 kmol fuel



15.88

Liquid butane at 25oC is mixed with 150% theoretical air at 600 K and is burned in a steady flow burner. Use the enthalpy of combustion from Table 15.3 to find the adiabatic flame temperature out of the burner. C4H10 + 1.5×6.5 (O2 + 3.76 N2 )→ 4 CO2 + 5 H2O + 3.25 O2 + 36.66 N2 Energy Eq.: Reactants:

HP - HR = 0

=>

°

°

°

∆HP = HR + ∆HR - HP = -HRP + ∆HR

∆HR = 9.75(9245) + 36.66(8894) = 416 193 kJ/kmol; ° HRP = M × HV = 58.124 × (-45 344) = -2 635 575 kJ/kmol -*

-*

-*

-*

∆HP = 4∆hCO2 + 5∆hH2O + 3.25 ∆hO2 + 36.66 ∆hN2 So the energy equation becomes

∆HP = 2 635 575 + 416 193 = 3 051 768 kJ/kmol Trial and Error:

LHS2000 K = 2 980 000,

Linear interpolation to match RHS

LHS2200 K = 3 369 866

=> TP = 2037 K



15.89

Gaseous ethanol, C2H5OH, is burned with pure oxygen in a constant volume combustion bomb. The reactants are charged in a stoichiometric ratio at the reference condition. Assume no heat transfer and find the final temperature ( > 5000 K). C2H5OH + 3 O2

→

2 CO2 + 3 H2O

Energy Eq.: - T = H°P + ∆H - n R -T UP = UR = H°R + ∆HR – nRR R P P P Solve for the properties that depends on T P and recall

∆HR = 0

° ° -0 -0 -0 ∆HP - nPRTP = HR - HP – nRRTR = hf fuel – 2 hf CO2 – 3 hf H2O - 4RTR

-0 Fuel: hf fuel = -235 000 kJ/kmol for IG from Table A.10 so ∆HP - nPRTP = -235 000 –2(-393 522) – 3(-241 826) – 4 × 8.31451 × 298.15 = 1 267 606 kJ/kmol LHS = ∆HP - nPRTP = 2 ∆hCO + 3 ∆hH O - 5 × 8.31451 × TP 2

2

From Table A.9 we find LHS5600 = 2 × 317 870 + 3 × 278 161 – 41.5726 × 5600 = 1 237 417 LHS6000 = 2 × 343 782 + 3 × 302 295 – 41.5726 × 6000 = 1 345 014 Tad.flame = 5712 K



15.90

The enthalpy of formation of magnesium oxide, MgO(s), is −601 827 kJ/kmol at 25°C. The melting point of magnesium oxide is approximately 3000 K, and the increase in enthalpy between 298 and 3000 K is 128 449 kJ/kmol. The enthalpy of sublimation at 3000 K is estimated at 418 000 kJ/kmol, and the specific heat of magnesium oxide vapor above 3000 K is estimated at 37.24 kJ/kmol K. a.

Determine the enthalpy of combustion per kilogram of magnesium.

b.

Estimate the adiabatic flame temperature when magnesium is burned with

a)

theoretical oxygen. 1 Mg + O2 → MgO(s) 2 -

∆hCOMB =

∆hCOMB M

=

h°f M

=

-601 827 = -24 746 kJ/kg 24.32

b) assume TR = 25°C and also that TP > 3000 K, 1st law: QCV = HP - HR = 0,

(MgO = vapor phase)

but HR = 0

⇒ HP = h°f + (h3000 - h298)SOL + ∆hSUB + CP VAP(TP - 3000) = -601 827 + 128 449 + 418 000 + 37.24(TP - 3000) = 0 Solving, TP = 4487 K



Second Law for the Combustion Process



15.91

Consider the combustion of hydrogen with pure oxygen in a stoichiometric ratio under steady flow adiabatic conditions. The reactants enter separately at 298 K, 100 kPa and the product(s) exit at a pressure of 100 kPa. What is the exit temperature and what is the irreversibility? The reaction equation is: H2 + νO2 O2

=> H2O

The balance of hydrogen is done, now for oxygen we - need vO2 = 0.5. Energy Eq.: HR = HP => 0 = -241 826 + ∆hH2O => ∆hH2O = 241 826 kJ/kmol Interpolate now in table A.9 for the temperature to give this enthalpy T = 4991 K

For this temperature we find from Table A.9, P = Po, so we do not need any pressure correction for the entropy ° ° SP = SP = s-H2O = 315.848 kJ/kmol K

For the reactants we have (again no pressure correction) ° ° SR = s-H2 + 0.5 s-O2 = 130.678 + 0.5 × 205.148 = 233.252 kJ/kmol K

S gen

= S – S = 315.848 – 233.252 = 82.596 kJ/kmol H K P

R

2

I = To Sgen = 298.15 × 82.596 = 24 626 kJ/kmol H2



15.92

Consider the combustion of methanol, CH3OH, with 25% excess air. The combustion products are passed through a heat exchanger and exit at 200 kPa, 400 K. Calculate the absolute entropy of the products exiting the heat exchanger assuming all the water is vapor. CH3OH + 1.25 × 1.5 (O2 + 3.76 N2) → CO2 + 2 H2O + 0.375 O2 + 7.05 N2 We need to find the mole fractions to do the partial pressures, n = 1 + 2 + 0.375 + 7.05 = 10.425 Gas mixture: ni

=>

yi = ni / n

yi

s-°i

- yi P -Rln P0

Si

CO2

1.0

0.0959

225.314

+13.730

239.044

H2O

2

0.1918

198.787

+7.967

206.754

O2

0.375

0.0360

213.873

+20.876

234.749

N2

7.05

0.6763

200.181

-2.511

197.670

SGAS MIX =

-

∑ n S = 2134.5 kJ/K kmol fuel i i



15.93

Two kilomoles of ammonia are burned in a steady flow process with x kmol of oxygen. The products, consisting of H2O, N2, and the excess O2, exit at 200°C, 7 MPa. a. Calculate x if half the water in the products is condensed. b. Calculate the absolute entropy of the products at the exit conditions. 2NH3 + xO2

→

3H2O + N2 + (x - 1.5)O2

Products at 200 oC, 7 MPa with a) yH2O VAP = PG/P =

nH2O LIQ = nH2O VAP = 1.5

1.5538 1.5 = 7 1.5 + 1 + x - 1.5

=>

x = 5.757

b) SPROD = SGAS MIX + SH2O LIQ Gas mixture:

ni

yi

s-°i

-Rln(yiP/P0)

S-i

H2O O2 N2

1.5 4.257 1.0

0.222 0.630 0.148

204.595 218.985 205.110

-22.810 -31.482 -19.439

181.785 187.503 185.671

SGAS MIX = 1.5(181.785) + 4.257(187.503) + 1.0(185.67) = 1256.55 kJ/K SH2O LIQ = 1.5[69.950 + 18.015(2.3223 - 0.3674)] = 157.75 kJ/K SPROD = 1256.55 + 157.75 = 1414.3 kJ/K



15.94

Propene, C3H6, is burned with air in a steady flow burner with reactants at P o, To. The mixture is lean so the adiabatic flame temperature is 1800 K. Find the entropy generation per kmol fuel neglecting all the partial pressure corrections. The reaction equation for a mixture with excess air is: C3H6 + νO2 (O2 + 3.76 N2)

→

3 H2O + 3 CO2 + 3.76νO2 N2 + (νO2 – 4.5)O2 °

Energy Eq.: The entropy equation:

°

°

HR = HR + ∆HR = HR = HP = HP + ∆HP ° SR + Sgen = SP => Sgen = SP – SR = SP – SR

From table A.9 at reference T

∆HR = ∆hFu + νO2(∆hO2 + 3.76 ∆hN2) = 0 From table A.9 at 1800 K:

∆HP = 3 ∆hH2O + 3 ∆hCO2 + 3.76 νO2 ∆hN2 + (νO2 – 4.5) ∆hO2 = 3 × 62 693 + 3 × 79432 + 3.76 νO2 × 48 979 + (νO2 – 4.5) 51 674 = 193 842 + 235 835 From table 15.3:

° HP

-

° HR

=

νO2

° HRP

= 42.081(–45 780) = –1 926 468 kJ/kmol

Now substitute all terms into the energy equation

Solve for νO2:

–1 926 468 + 193 842 + 235 835 νO2 = 0 1 926 468 – 193 842 = 7.3468, νN2 = 27.624 νO2 = 235 835

Table A.9-10 contains the entropies at 100 kPa so we get: SP = 3 × 259.452 + 3 × 302.969 + (7.3468 – 4.5) 264.797 + 27.624

× 248.304

= 9300.24 kJ/kmol-K SR = 267.066 + 7.3468

× 205.148 + 27.624 × 191.609 = 7067.25 kJ/kmol K

Sgen = 9300.24 – 7067.25 = 2233 kJ/kmol-K



15.95

A flow of hydrogen gas is mixed with a flow of oxygen in a stoichiometric ratio, both at 298 K and 50 kPa. The mixture burns without any heat transfer in complete combustion. Find the adiabatic flame temperature and the amount of entropy generated per kmole hydrogen in the process. The reaction equation is: H2 + νO2 O2

=> H2O

The balance of hydrogen is done, now for oxygen we - need vO2 = 0.5. Energy Eq.: HR = HP => 0 = -241 826 + ∆hH2O => ∆hH2O = 241 826 kJ/kmol Interpolate now in table A.9 for the temperature to give this enthalpy T = 4991 K

For this temperature we find from Table A.9 ° SP = s-H2O – R ln(P/Po) = 315.848 – 8.31451 ln(0.5) = 321.611 kJ/kmol K

For the reactants we have ° ° SR = s-H2 – R ln(P/Po) + 0.5 [s- O2 – R ln(P/Po) ]

= 130.678 + 0.5 × 205.148 - 1.5 × 8.31451 ln(0.5) = 241.897 kJ/kmol K Sgen = SP – SR = 321.611 – 241.897 = 79.714 kJ/kmol H2 K Recall that this includes the mixing process.



15.96

Calculate the irreversibility for the process described in Problem 15.45. 2 C + 2 O2

→ 1 CO2 + 1 CO + 12 O2

Process

V = constant,

C: solid, n1(GAS) = 2, n2(GAS) = 2.5

P2 = P1 ×

n2T2 2.5 × 1000 = 200 × = 838.4 kPa n1T1 2 × 298.2

H1 = HR = 0 H2 = HP = 1(-393522 + 33397) + 1(-110527 + 21686) + (1/2)(0 + 22703) = -437 615 kJ Q = (U U ) = (H2 - H1) - n2RT2 + n1RT1 1 2 2 1 = (-437 615 - 0) - 8.3145(2.5 × 1000 - 2 × 298.2) = -453 442 kJ

COMBUSTION GAS cb

Reactants: 200 SR = 2(5.740) + 2(205.148 - 8.31451 ln ) = 410.250 kJ/K 100 Products: ni

yi

s-°i

- yi P -Rln P0

S-i

CO2

1.0

0.40

269.299

-10.061

259.238

CO O2

1.0 0.5

0.40 0.20

234.538 243.579

-10.061 -4.298

224.477 239.281

SP = 1.0(259.238) + 1.0(224.477) + 0.5(239.281) = 603.355 kJ/K I = T0(SP - SR) - 1Q2 = 298.15(603.355 - 410.250) - (-453 442) = +511 016 kJ



15.97

Consider the combustion of methanol, CH3OH, with 25% excess air. The combustion products are passed through a heat exchanger and exit at 200 kPa, 40°C. Calculate the absolute entropy of the products exiting the heat exchanger per kilomole of methanol burned, using the proper amounts of liquid and vapor for the water. CH3OH + 1.25 × 1.5 (O2 + 3.76 N2) → CO2 + 2 H2O + 0.375 O2 + 7.05 N2 o

Products exit at 40 C, 200 kPa, check for saturation: PG 7.384 nV MAX yV MAX = = = P 200 nV MAX + 1 + 0.375 + 7.05 nV = nV MAX = 0.323 Gas mixture: ni

nLIQ = 1.677

yi

s-°i

- yi P -Rln P0

Si

CO2

1.0

0.1143

215.633

+12.270

227.903

H2O

0.323

0.0369

190.485

+21.671

212.156

O2

0.375

0.0429

206.592

+20.418

227.01

N2

7.05

0.8059

193.039

-3.969

189.07

SGAS MIX =

∑ n S- = 1714.50 kJ/K kmol fuel i i

s-LIQ = 69.950 + 18.015(0.5725 - 0.3674) = 73.645 kJ/kmol SLIQ = 1.677 × 73.645 = 123.50 kJ/K kmol fuel SPROD = 1714.50 + 123.50 = 1838 kJ/K kmol fuel



15.98

Graphite, C, at P0, T0 is burned with air coming in at P0, 500 K in a ratio so the products exit at P0, 1200 K. Find the equivalence ratio, the percent theoretical air, and the total irreversibility. C + (1/φ)b(O2 + 3.76 N2)

→

CO2 + ((1/φ) - 1 )O 2 + 3.76 (1/φ) N2

Stoichiometric: b = 1 Energy Eq.:

°

°

HP = HR ⇒ ∆HP 1200 - ∆HR = HR - HP

44 473 + ((1/φ) - 1 )29 761 + 3.76(1/ φ)28 109 - (1/φ)(6086 + 3.76×5911) = 0 - (-393 522)

Sgen = SP - SR =

⇒

(1/φ) = 3.536

∑ν(s- - R- ln(y)) °

P-R

Reactants:

yO = 0.21, yN = 0.79

Products:

yO = 0.1507,

2

2

2

(carbon is solid)

yN = 0.79, 2

yCO = 0.0593 2

° SP

= 279.39 + 2.536 × 250.011 + 13.295 × 234.227 = 4027.5

° SR

= 5.74 + 3.536(220.693 + 3.76

× 206.74) = 3534.8

For the pressure correction the term with the nitrogen drops out (same y). R

∑-ν ln(y) = R- [-ln(yCO2) - 2.536 ln(yO2 prod) + 3.536 ln(yO2 reac )]

P-R

= R [ -ln(0.0593) - 2.536 ln(0.1507) + 3.536 ln(0.21) ] = 8.3145(2.8235 + 2.536 ×1.8927 - 3.536 ×1.5606) = 17.503

Sgen = 4027.5 - 3534.8 + 17.503 = 510.2 kJ/kmol carbon-K kJ

I = T0 Sgen = 152 117 kmol C



15.99

An inventor claims to have built a device that will take 0.001 kg/s of water from the faucet at 10°C, 100 kPa, and produce separate streams of hydrogen and oxygen gas, each at 400 K, 175 kPa. It is stated that this device operates in a 25 °C room on 10-kW electrical power input. How do you evaluate this claim? Liq H2O

H2 gas

o

O2 gas

10 C, 100 kPa 0.001 kg/s

. -WCV = 10 kW

 each at  400 K  175 kPa T0 = 25 oC

1

H2O → H2 + 2 O2 1

Hi - He = [-285830 + 18.015(42.01 - 104.89)] - 2961 - 2 (3027) = -291 437 kJ/kmol (Si - Se) = [69.950 + 18.015(0.151 - 0.3674)] - (139.219 - 8.3145 ln 1.75) 1

- 2 (213.873 - 8.3145 ln 1.75) = -173.124 kJ/kmol K WREV = (Hi - He) - T0(Si - Se) = -291 437 - 298.15(-173.124) = -239820 kJ/kmol . WREV = (0.001/18.015)(-239 820) = -13.31 kW . . . I = WREV - WCV = -13.31 - (-10) < 0

Impossible



15.100

Hydrogen peroxide, H2O2, enters a gas generator at 25 °C, 500 kPa at the rate of 0.1 kg/s and is decomposed to steam and oxygen exiting at 800 K, 500 kPa. The resulting mixture is expanded through a turbine to atmospheric pressure, 100 kPa, as shown in Fig. P15.100. Determine the power output of the turbine, and the heat transfer rate in the gas generator. The enthalpy of formation of liquid H 2O2 is

−187 583 kJ/kmol.

. mH2O2 1 0.1 . H2O2 → H2O + 2 O2 nH2O2 = M = 34.015 = 0.00294 kmol/s . . nMIX = nH2O2× 1.5 = 0.00441 kmol/s

2 1 CP0 MIX = 3 × 1.872 × 18.015 + 3 × 0.922 × 31.999 = 32.317 CV0 MIX = 32.317 - 8.3145 = 24.0 => kMIX = 32.317/24.0 = 1.3464 CV: turbine. Assume reversible → s3 = s2 P3 k-1 100 0.2573 T3 = T2( ) k = 800( ) = 528.8 K P2 500 w = CP0(T2 - T3) = 32.317(800 - 528.8) = 8765 kJ/kmol . WCV = 0.00441 × 8765 = 38.66 kW CV: Gas Generator . H . 1 = 0.00294(-187 583 + 0) = -551.49 H2 = 0.00294(-241 826 + 18002) + 0.00147(0 + 15836) = -634.76 . . . QCV = H2 - H1 = -634.76 + 551.49 = -83.27 kW



15.101

Methane is burned with air, both of which are supplied at the reference conditions. There is enough excess air to give a flame temperature of 1800 K. What are the percent theoretical air and the irreversibility in the process? The combustion equation with X times theoretical air is CH4 + 2X(O2 + 3.76 N2)  CO2 + 2H2O + 2(X-1)O2 + 7.52X N2 Energy Eq.:

H

°

fuel ° HP = HR

⇒

∆

From Table 15.3:

°

= H = H = H + ∆H = H + ∆H

+H air

R

P

P

°

°

P

R

R

+ ∆HR - HP = -HRP + 0

°

-HRP = -16.04 (-50 010) = 802 160 kJ/kmol -*

-*

-*

-*

∆HP = ∆hCO2 + 2 ∆hH2O + 2(X-1) ∆hO2 + 7.52X ∆hN2 From Table A.9 and the energy equation

∆HP 1800 = 79 432 + 2 × 62 693 + 2(X-1) 51 674 + 7.52X × 48 979 = 802 160 so 101 470 + 471 670 X = 802 160 => %Theoretical air = 148.6% The products are Products:

X = 1.4856

CO2 + 2H2O + 0.9712 O2 + 11.172 N2

The second law S

=S -S

gen

P

and

I=T S

R

o

gen

Reactants: Pi = 100 kPa, Po = 100 kPa, − s f from Table A.9 o

ni

yi

−s o f

−

-R ln

yiPi Po

Si

kJ kmol K

CH4

1

1

186.251

0

186.251

O2

2X

0.21

205.148

12.976

218.124

N2

7.52 X

0.79

191.609

1.96

193.569

SR =

-

∑ niSi = 2996.84 kJ/K kmol fuel



Products: Pe = 100 kPa, Po = 100 kPa ni

yi

−s o

1800

−

- R ln

yiPe Po

Si

kJ kmol K

CO2

1

0.06604 302.969

22.595

325.564

H 2O

2

0.13208 259.452

16.831

276.283

O2

0.9712

0.06413 264.797

22.838

287.635

N2

11.172

0.73775 248.304

2.529

250.833

SP =

-

∑ niSi = 3959.72 kJ/K kmol fuel;

I = To(SP - SR) = 298.15(3959.72 - 2996.84) = 287 MJ/kmol fuel



15.102

Pentane gas at 25°C, 150 kPa enters an insulated steady flow combustion chamber. Sufficient excess air to hold the combustion products temperature to 1800 K enters separately at 500 K, 150 kPa. Calculate the percent theoretical air required and the irreversibility of the process per kmol of pentane burned. C5H12 + 8X( O2 + 3.76 N2)  5 CO2 + 6 H2O + 8(X-1) O2 + 30.08X N2

Energy Eq.:

Qcv + HR = HP + WCV; WCV= 0, Qcv = 0

−o

− ∆h500 for O2 and N2 −o − − HR = ( hf )C5H12 + 8X ∆hO2 + 30.08X ∆hN2

Reactants: C5H12 : hf from A.9

and

from A.9

= -146 500 + 8X 6086 + 30.08 X 5911 = 226 491 X - 146 500

−o

−

−o

−

−

−

HP = 5( hf + ∆h )CO2 + 6( hf + ∆ h)H2O + 8(X-1) ∆hO2 + 30.08 X ∆hN2 = 5(-393 522 + 79 432) + 6(-241 826 + 62 693) + 8(X-1) 51 674 + 30.08 X 48 979 = 1 886 680 X - 3 058 640 Energy Eq. solve for X; HR = HP = 226 491 X - 146 500 = 1 886 680 X - 3 058 640

⇒

X = 1.754

b) Reactants: Pin = 150 kPa, Po = 100 kPa, − sf

o

ni

yi

−s o, −s o

- R ln

f

500

−

yiPin Po

Si

kJ kmol K

C5H12

1

1

348.945

-3.371

345.574

O2

8X

0.21

220.693

9.605

230.298

N2

30.08 X

0.79

206.74

-1.411

205.329

SR =

-

∑ niSi = 14410.34 kJ/K kmol fuel



Products: Pe = 150 kPa, Po = 100 kPa

ni

yi

−s o

1800

- R ln

−

yiPe Po

Si

kJ kmol K

CO2

5

0.0716

302.969

18.550

321.519

H 2O

6

0.086

259.452

17.027

276.479

O2

8(X-1)

0.0864

264.797

16.988

281.785

N2

30.08X

0.756

248.304

-1.045

247.259

SP =

-

∑ niSi = 17 732.073

kJ/K kmol fuel;

I = To(SP - SR) = 298.15(17 732.07 - 14 410.34) = 990 MJ/kmol fuel



15.103

A closed rigid container is charged with propene, C 3H6, and 150% theoretical air at 100 kPa, 298 K. The mixture is ignited and burns with complete combustion. Heat is transferred to a reservoir at 500 K so the final temperature of the products is 700 K. Find the final pressure, the heat transfer per kmole fuel and the total entropy generated per kmol fuel in the process. C H +ν 3

6

→

O + 3.76 N O2

(

2

)

Oxygen, O2 , balance: Actual Combustion:

2

npT2 nRT1

2

2 νO = 6 + 3 = 9 2

φ = 1.5 ⇒ νO

C3H6 + 6.75 O2 + 25.38 N2 P2 = P1

3 CO + 3 H O + x N

2

= 100 ×

→

2 ac

2

⇒

νO2= 4.5

= 1.5 × 4.5 = 6.75

3 CO2 + 3 H2O + 25.38 N2 + 2.25 O2

33.63 × 700 = 238.3 kPa 33.13 × 298.15

Enthalpies from Table A.9

∆HP 700 = 3×17 754 + 3×14 190 + 25.38×11 937 + 2.25×12 499 kJ

= 426 916 kmol fuel Enthalpy of combustion from table 15.3 converted to mole basis °

HRP = -45 780 × 42.081 = -1 926 468 kJ/kmol fuel U2 - U1 = 1Q2 - 0 = H2 - H1 - n2R  T2 + n1R  T1 °

Q2 = HRP + ∆HP 700 - nPR  T2 + n1R  T1

1

= -1 926 468 + 426 916 - 33.63

× 8.3145 × 700 kJ

+ 33.13 × 8.3145 × 298.15 = -1.613 106 kmol fuel Entropies from Table A.9 and pressure correction ni

yi

s-°i

-Rln(yiP/P0)

C3H8

1.0

0.0302

267.066

29.104

296.17

O2

6.75

0.2037

205.143

13.228

218.376

N

25.38 0.7661

191.609

2.216

189.393

Reactants:

S-i

2


Borgnakke and Sonntag kJ

S1= 296.17 + 6.75 × 218.376 + 25.38 × 189.393 = 6577 kmol fuel K

Products:

ni

s-°i

yi

-Rln(yiP/P0)

S-i

CO2

3

0.0892

250.752

+12.875

263.627

H2O

3

0.0892

218.739

+12.875

231.614

O2

2.25

0.0669

231.465

+15.266

246.731

N

25.38

0.7547

216.865

-4.88

211.985

2

S2 = 3(263.627 + 231.614) + 2.25 × 246.731 + 25.38 × 211.985 = 7421 kJ/kmol fuel K S

1 2 gen

= S2 - S1 - 1Q2/Tres = 7421 - 6577 +

1.613×106 kJ = 4070 kmol fuel K 500



Problems Involving Generalized Charts or Real Mixtures



15.104

A gas mixture of 50% ethane and 50% propane by volume enters a combustion chamber at 350 K, 10 MPa. Determine the enthalpy per kilomole of this mixture relative to the thermochemical base of enthalpy using Kay’s rule. -* hMIX O = 0.5(-84740) + 0.5(-103900) = -94320 kJ/kmol CP0 MIX = 0.5 × 30.07 × 1.7662 + 0.5 × 44.097 × 1.67 = 63.583 -* -* h350 - h298 = 63.583(350 - 298.2) = 3294 kJ/kmol Kay’s rule:

TC MIX = 0.5 × 305.4 + 0.5 × 369.8 = 337.6 K PC MIX = 0.5 × 4.88 + 0.5 × 4.25 = 4.565 MPa

Tr = 350/337.6 = 1.037,

Pr = 10/4.565 = 2.19

-* From Fig. D.2: h - h = 8.3145 × 337.6 × 3.53 = 9909 kJ/kmol hMIX 350K,10MPa = -94320 + 3294 - 9909 = -100 935 kJ/kmol



15.105

Liquid butane at 25°C is mixed with 150% theoretical air at 600 K and is burned in an adiabatic steady state combustor. Use the generalized charts for the liquid fuel and find the temperature of the products exiting the combustor. o

25 C LIQ. C4 H10 150% Air 600 K

Prod.

Adiab. Comb.

at T p

TC = 425.2 K TR = 0.701

QCV = 0

-o hC4H10 = hf IG + (hLIQ - h*)

see Fig. D.2

= -126200 + (-4.85 × 8.3145 × 425.2) = -143 346 kJ C4H10 + 1.5 × 6.5 O2 + 3.76 × 9.75 N2

→ 4 CO2 + 5 H2O + 3.25 O2 + 36.66 N2 hAIR = 9.75(9245) + 36.66(8894) = 416 193 kJ ⇒

HR = 416 193- 143 346 = +272 847 kJ

-* -* -* -* HP = 4(-393522 + ∆hCO2) + 5(-241826 + ∆hH2O) + 3.25 ∆hO2 + 36.66 ∆hN2 Energy Eq.:

HP - HR = 0

-* -* -* -* 4 ∆hCO2 + 5 ∆hH2O + 3.25 ∆hO2 + 36.66 ∆hN2 = 3 056 065 Trial and Error: LHS2000 K = 2 980 000, LHS2200 K = 3 369 866 Linear interpolation to match RHS

=> TP = 2039 K



15.106

Repeat Problem 15.135, but assume that saturated-liquid oxygen at 90 K is used instead of 25°C oxygen gas in the combustion process. Use the generalized charts to determine the properties of liquid oxygen. Problem same as 15.135, except oxygen enters at 2 as saturated liquid at 90 K. . . . . . . mO2/mN2H4 = 0.5 = 32nO2/32nN2H4 and Q/mN2H4 = -100 kJ/kg Energy Eq.:

QCV = HP - HR = -100 × 32.045 = -3205 kJ/kmol fuel 1

Reaction equation:

1 N2H4 + 2 O2

At 90 K, Tr2 = 90/154.6 = 0.582

→

⇒

H2O + H2 + N2

~ ∆hf = 5.2

-* (h - h) = 8.3145 × 154.6 × 5.2 = 6684 kJ/kmol

Figure D.2, ∆hAT 2 = -6684 + 0.922 × 32(90 - 298.15) = -12825 kJ/kmol 1

HR = 50417 + 2(0 - 12825) = 44005 kJ, 1st law: ∆hP = ∆hH

2O

From Table A.9,

H°P = -241826

+ ∆hH + ∆hN = Qcv + HR - HP° = 282626 2

2

∆HP 2800K = 282141, ∆HP 3000K = 307988

Therefore, TP = 2804 K



15.107

A mixture of 80% ethane and 20% methane on a mole basis is throttled from 10 MPa, 65°C, to 100 kPa and is fed to a combustion chamber where it undergoes complete combustion with air, which enters at 100 kPa, 600 K. The amount of air is such that the products of combustion exit at 100 kPa, 1200 K. Assume that the combustion process is adiabatic and that all components behave as ideal gases except the fuel mixture, which behaves according to the generalized charts, with Kay’s rule for the pseudocritical constants. Determine the percentage of theoretical air used in the process and the dew-point temperature of the products. Reaction equation: -0 Fuel mix: hf FUEL = 0.2(-74873) + 0.8(-84740) = -82767 kJ/kmol CP0 FUEL = 0.2 × 2.2537 × 16.04 + 0.8 × 1.7662 × 30.07 = 49.718 -*

∆hFUEL = 49.718(65 - 25) = 1989 kJ/kmol TCA = 305.4 K, TCB = 190.4 K PCA = 4.88, PCB=4.60

⇒

⇒

Tc mix=282.4 K

Pc mix= 4.824 MPa

Tr = 338.2/282.4 = 1.198, Pr = 10/4.824 = 2.073 (h* - h)FUEL IN = 8.31451 × 282.4 × 2.18 = 5119 -

kJ

⇒ hFUEL IN = -82767 + 1989 - 5119 = -85897 kmol Energy Eq.: 1.8(-393522 + 44473) + 2.8(-241826 + 34506) + 3.2(x - 1)(29761) + (12.03x)(28109) + 85897 - (3.2x)(9245) - (12.03x)(8894) = 0 a)

x = 4.104 or 410.4 %

b) nP = 1.8 + 2.8 + 3.2(4.104 - 1) + 12.03 yH2O = 2.8/63.904 = PV/100 ;

× 4.104

= 63.904

PV = 4.38 kPa, T = 30.5°C



15.108

Saturated liquid butane enters an insulated constant pressure combustion chamber at 25°C, and x times theoretical oxygen gas enters at the same P and T. The combustion products exit at 3400 K. With complete combustion find x. What is the pressure at the chamber exit? and what is the irreversibility of the process? Butane: T1 = To = 25oC, sat liq., x1 = 0, Tc = 425.2 K, Pc = 3.8 MPa Do the properties from the generalized charts Fig. D.1: Tr1 = 0.7, Pr1 = 0.1, P1 = Pr1Pc = 380 kPa

− −h* − −h  = 4.85 R −* - −s 1)f = Tc , (s  1 1 f 1

Figs. D.2 and D.3:

Oxygen: T2 = To = 25oC,

−

6.8 R

X - Theoretical O2

Products: T3 = 3400 K, Assumes complete combustion C4H10 + 6.5X O2 Energy Eq.:



4 CO2 + 5 H2O + 6.5(X-1) O2

Qcv + HR = HP + Wcv; Qcv = 0, Wcv = 0

−o

−

HR = n( hf + ∆ h)C4H10 = 1(-126 200 + -17 146) = -143 346 kJ

−o

−

Products: CO2

n( hf + ∆h )CO2 = 4(-393 522 + 177 836) = -862 744 kJ

H2O

n(− hof + ∆− h )H2O = 5(-241 826 + 149 073) = -463 765 kJ

O2

n( hf + ∆h )O2 = 6.5(X-1)(0 + 114 101) = (X-1)741 657 kJ

−o

HP = Energy Eq.:

−

−o

−

∑ni (hf + ∆h)i = 741 657X – 2 068 166 HP = HR solve for X;

X = 2.594

Assume that the exit pressure equals the inlet pressure:

−s C

4H10

−

= [− s f - R ln o

P1 Po

−

− - −s 1)f] ; −s O = [−s o - R ln - (s 1 *

2

Pe = Pi = 380 kPa P1 Po

]

SR = SC4H10 + SO2 = [306.647 - 11.10 - 56.539] + [205.48 - 11.10] × 6.5 × 2.594 = 3516.45 kJ/K Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.


Products: ni

yi

−s o

− yiPe − R ln P

Si

CO2

4

0.2065

341.988

2.016

344.004

H 2O

5

0.2582

293.550

0.158

293.708

O2

10.368

0.5353

289.499

-5.904

283.595

i

o

SP =

kJ kmol K

-

∑ niSi = 5784.87 kJ/K;

I = To(SP - SR) = 298.15 (5784.87 - 3516.45) = 676 329 kJ



15.109

Liquid hexane enters a combustion chamber at 31 °C, 200 kPa, at the rate 1 kmol/s 200% theoretical air enters separately at 500 K, 200 kPa, and the combustion products exit at 1000 K, 200 kPa. The specific heat of ideal gas hexane is C = p 143 kJ/kmol K. Calculate the rate of irreversibility of the process. Hexane:

Tc = 507.5 K, Pc = 3010 kPa

Tr1 = 0.6, Fig. D.1: Prg = 0.028, Pg1 = Pr1Pc = 84.47 kPa

−* −

−

−

*

Figs D.2 and D.3: (h 1 - h 1) f = 5.16 RTc , (s −1 - −s 1) f = 8.56 R Air: T2 = 500 K, P2 = 200 kPa, 200% theoretical air Products: T3 = 1000 K, P3 = 200 kPa

−

−o

−* −

−* −*

−* −

a) h C6H14 = h f - (h 1 - h 1)f + (h 1 - h 0) + (h 0 - h 0)

−* −

−* −* −

−o

h0 - h 0 = 0 , h1 - h 0 = CP (T1 - To) = 858 kJ/kmol, h f = -167300 kJ/kmol

−* −

−

h1 - h 1 = 5.16 × 8.3145 × 507.5 = 21773 kJ/kmol, hC6H14 = -188215 kJ/kmol T1 − P1 − − −* To + CP ln T - R ln P + (s 1 - s 1) o o

−s C6H14 = −s o

T1 − P1 − To + CP ln T - R ln P = 387.979 + 2.85 – 5.763 = 385.066 kJ/kmol-K o o

−s o *

−s 1 - −s 1 = 8.56×8.3145 = 71.172 kJ/kmol-K, −s C6H14 = 313.894 kJ/kmol-K b) C6H14 + 19O2 + 71.44N2  6CO2 + 7H2O + 9.5O2 + 71.44N2 T3 Tc prod = yiTci = 179.3 K, Tr3 = = 5.58 Ideal Gas Tc prod

∑

c) 1st Law: Q + HR = HP + W;

−

W=0

−

=>

Q = HP - HR

−

HR = ( h)C6H14 + 19∆h O2 + 71.44 ∆ hN2 = -188 215 + 19 6086 + 71.44 5911 = 349701 kJ/kmol fuel

−o

−

−o

−

−o

−

−o

−

HP = 6(h f + ∆h )CO2 = 7(h f + ∆h )H2O + 9.5 (h f + ∆h )O2 + 71.44(h f + ∆h)N2 CO2

-

(−hof + ∆−h) = (-393522 + 33397) = -360125 kJ/kmol

H2O

-

(−hof + ∆−h) = (-241826 + 26000) = -215826 kJ/kmol

O2

-

(−hof + ∆−h) = (0 + 22703) = 22703 kJ/kmol



N2

(−hof + ∆−h) = (0 + 21463) = 21463 kJ/kmol

-

. HP = -1922537 kJ; Q = -2272238 kW . . . d) I = To n (SP - SR) - Q; To = 25oC

−

SR = (− s )C6H14 + 19 (s−500 - Rln o

yO2P2 Po

−

)O2 + 71.44(s−500 - Rln o

yN2P2 Po

)N2

(−s )C6H14 = 313.894 kJ/kmol K, (−s 500)O2 = 220.693 kJ/kmol K o

o

(. −s 500)N2 = 206.740 kJ/kmol K,

yO2 = 0.21, yN2 = 0.79

SR = 19141.9 kW/K Products: ni

yi

−s o i

− yiPe − R ln P o

Si (kJ/kmol-K)

CO2

6

0.0639

269.299

17.105

286.404

H 2O

7

0.0745

232.739

15.829

248.568

O2

9.5

0.1011

243.579

13.291

256.87

N2

71.44

0.7605

228.171

-3.487

224.684

SP = ni− s i = 21950.1 kJ/K; . . . I = T n (S - S ) - Q = 3 109 628 kW P R o

∑



Fuel Cells



15.110

In Example 15.16, a basic hydrogen–oxygen fuel cell reaction was analyzed at 25°C, 100 kPa. Repeat this calculation, assuming that the fuel cell operates on air at 25°C, 100 kPa, instead of on pure oxygen at this state. Anode: 2 H2

→ 4 e - + 4 H+

Cathode: 4 H+ + 4 e- → 1 O2 + 2 H2O Overall: 2 H2 + 1 O2 → 2 H2O Example 14.16:

∆G25°C = -474 283 kJ (for pure O2)

For PO2 = 0.21 × 0.1 MPa: SO = 205.148 – 8.3145 ln 0.21 = 218.124 kJ/kmol 2

∆S = 2(69.950) – 2(130.678) – 1(218.124) = -339.58 kJ/kmol K ∆G25°C = -571 660 – 298.15(-339.58) = -470 414 kJ/kmol E° =

470414 = 1.219 V 96487 × 4



15.111

Assume that the basic hydrogen-oxygen fuel cell operates at 600 K instead of 298 K as in example 15.16. Find the change in the Gibbs function and the reversible EMF it can generate.

Reaction:

2 H2 + O2

⇒

2 H 2O

At a temperature of 600 K the water formed is in a vapor state. We can thus find the change in the enthalpy as

−o − −o − −o − 0 ∆H600 K = 2(hf + ∆h)H2O g - 2(hf + ∆h)H2 - (hf + ∆h)O2 = 2(-241 826 + 10 499) – 2(0 + 8799) – 0 – 9245 = -489 497 kJ/4 kmol e0 S600 K

∆

−o

−o

−o

= 2 s f H2O g – 2 s f H2 – s f O2 = 2 × 213.051 – 2 × 151.078 – 226.45 = -102.504 kJ/4 kmol e- K

0 G600 K

∆

0

0

= ∆H600 K – T∆S600 K = -489 497 – 600(-102.504)

= - 427 995 kJ/4 kmol erev W = -∆G0 = 427 995 kJ/4 kmol e-∆G0 427 995 E0 = = = 1.109 V 96485 × 8 96 485 × 4



15.112

For a PEC fuel cell operating at 350 K the constants in Eq.15.29 are: ileak = 0.01, iL = 2, io = 0.013 all A/cm2, b = 0.08 V, c = 0.1 V, ASR = 0.01 Ω cm2 and EMF = 1.22 V. Find the voltage and the power density for the current density i = 0.25, 0.75 and 1.0 A/cm 2. The fuel cell output voltage is from Eq.15.29: i + ileak iL V = EMF – b ln ( ) – i ASR – c ln ( ) ohmic io iL – (i + ileak) i = 0.25:

V = 1.22 – 0.23966 – 0.0025 – 0.01393 = 0.9639 V, p = Vi = 0.241 W/cm2

i = 0.75:

V = 1.22 – 0.32547 – 0.0075 – 0.0478 = 0.8392 V, p = Vi = 0.6294 W/cm2

i = 1.0:

V = 1.22 – 0.34822 – 0.01 – 0.0703 = 0.7915 V, p = Vi = 0.7915 W/cm2



15.113

Assume the PEC fuel cell in the previous problem. How large an area does the fuel cell have to deliver 1 kW with a current density of 1 A/ cm2? The fuel cell output voltage is from Eq.15.29: i + ileak iL V = EMF – b ln ( ) – i ASRohmic – c ln ( ) io iL – (i + ileak) i = 1.0:

V = 1.22 – 0.34822 – 0.01 – 0.0703 = 0.7915 V, p = Vi = 0.7915 W/cm2

Total power

P = p A = 0.7915 W/cm2

× A = 1000 W

A = (1000 / 0.7915) cm2 = 1263 cm2



15.114

Consider a methane-oxygen fuel cell in which the reaction at the anode is CH4 + 2H2O

→

CO2 + 8e- + 8H+

The electrons produced by the reaction flow through the external load, and the positive ions migrate through the electrolyte to the cathode, where the reaction is 8 e - + 8 H + + 2 O 2 → 4 H 2O Calculate the reversible work and the reversible EMF for the fuel cell operating at 25°C, 100 kPa. CH4 + 2H2O → CO2 + 8e- + 8H+ and 8e - + 8H+ + 2CO2 → 4H2O Overall

CH4 + 2O2 → CO2 + 2H2O

o

a) 25 C assume all liquid H2O and all comp. at 100 kPa 0

∆H25 C = -393 522 + 2(-285 830) – (-74 873) – 0 = -890 309 kJ 0

∆S25 C = 213.795 + 2(69.950) – 186.251 – 2(205.148) = - 242.852 kJ/K 0

∆G25 C = -890 309 – 298.15(-242.852) = - 817 903 kJ Wrev = -∆G0 = +817903 kJ -∆G0 +817903 0 E = 96485 × 8 = 96485 × 8 = 1.06 V



15.115

Consider a methane-oxygen fuel cell in which the reaction at the anode is CH4 + 2H2O

→

CO2 + 8e- + 8H+

The electrons produced by the reaction flow through the external load, and the positive ions migrate through the electrolyte to the cathode, where the reaction is 8 e - + 8 H + + 2 O 2 → 4 H 2O Assume that the fuel cell operates at 1200 K instead of at room temperature. CH4 + 2H2O → CO2 + 8e- + 8H+ and 8e- + 8H+ + 2CO2 → 4H2O CH4 + 2O2 → CO2 + 2H2O

Overall

At 1200 K assume all gas H 2O and all comp. at 100 kPa 0

∆H1200 K = 1(-393522 + 44473) + 2(-241826 + 34506)

- 2(0 + 29761)

- 1[-74873 + 16.043 × 2.254(1200 - 298.2)] = -780 948 kJ 0

∆S1200 K = 1(279.390) + 2(240.485) - 1(186.251 + 16.043 × 2.254 ln

1200 ) - 2(250.011) 298.2

= 23.7397 kJ/K 0 G1200 K

∆

0

0

= ∆H1200 K - T∆S1200 K = -780 948 - 1200(23.7397) = -809 436 kJ

W

rev

= +809 436 kJ

E0=

+809 436 = 1.049 V 96 485 × 8



15.116

A SOC fuel cell at 900 K can be described by an EMF = 1.06 V and the constants in Eq.15.29 as : b = 0 V, c = 0.1 V, ASR = 0.04 Ω cm2, ileak = 0.01, iL = 2, io = 0.13 all A/cm2. Find the voltage and the power density for the current density i = 0.25, 0.75 and 1.0 A/cm 2. The fuel cell output voltage is from Eq.15.29: i + ileak iL V = EMF – b ln ( ) – i ASR – c ln ( ) ohmic io iL – (i + ileak) i = 0.25:

V = 1.06 – 0.0 – 0.01 – 0.01393 = 1.036 V, p = Vi = 0.259 W/cm2

i = 0.75:

V = 1.06 – 0.0 – 0.03 – 0.0478 = 0.9822 V, p = Vi = 0.737 W/cm2

i = 1.0:

V = 1.06 – 0.0 – 0.04 – 0.0703 = 0.95 V, p = Vi = 0.95 W/cm2



15.117

Assume the SOC fuel cell in the previous problem. How large an area does the fuel cell have to deliver 1 kW with a current density of 1 A/ cm 2? The fuel cell output voltage is from Eq.15.29: i + ileak iL V = EMF – b ln ( ) – i ASRohmic – c ln ( ) io iL – (i + ileak) i = 1.0:

V = 1.06 – 0.0 – 0.04 – 0.0703 = 0.95 V, 2

p = Vi = 0.95 W/cm Total power

P = p A = 0.95 W/cm2

× A = 1000 W

2

A = (1000 / 0.95) cm = 1053 cm2



15.118

A PEC fuel cell operating at 25oC, generates 1.0 V that also account for losses. For a total power of 1 kW what is the hydrogen mass flow rate? The work term for the fuel cell is 0

W = - ∆G298 K . . m = Mn =

. MW 0

-∆G298 K

so =

. . 0 W = -n ∆G298 K

2.016 kg/kmol × 1 kW -6 = 8.5 × 10 kg/s 0.5 × 474 283 kJ/kmol

Since the actual output was 1 V rather than the ideal 1.229 V we then infer an efficiency of (1/1.229 = 0.814) so . . -6 mac = m / 0.814 = 10.45 10 kg/s



15.119

A basic hydrogen-oxygen fuel cell operates at 600 K instead of 298 K as in example 15.16. For a total power of 5 kW, find the hydrogen mass flow rate and the exergy in the exhaust flow. Reaction:

2 H2 + O2

⇒

2 H 2O

At a temperature of 600 K the water formed is in a vapor state. We can thus find the change in the enthalpy as

∆H0600 K = 2(−hof + ∆−h)H2O g - 2(−hof + ∆−h)H2 - (−hof + ∆−h)O2 = 2(-241 826 + 10 499) – 2(0 + 8799) – 0 – 9245 = -489 497 kJ/4 kmol e0 S600 K

∆

−o

−o

−o

= 2 s f H2O g – 2 s f H2 – s f O2 = 2 × 213.051 – 2 × 151.078 – 226.45 = -102.504 kJ/4 kmol e- K

0 G600 K

∆

0

0

= ∆H600 K – T∆S600 K = -489 497 – 600(-102.504)

= - 427 995 kJ/4 kmol erev W = -∆G0 = 427 995 kJ/4 kmol eThis was for 2 kmol hydrogen . MW 2.016 kg/kmol × 5 kW . . 0 m = Mn = = 0.5 × 427 995 kJ/kmol = 47.1 -∆G

-6

10

kg/s

600 K

The exhaust flow is water vapor at 600 K, 100 kPa so use Table B.1 . . mψ = m [(h – ho – To(s – so)] -6

= 47.1 × 10 [3129 – 104.87 – 298.15 (8.3037 – 0.3673)] = 0.031 kW



Combustion applications and efficiency



15.120

For the combustion of methane 150% theoretical air is used at 25 oC, 100 kPa and relative humidity of 70%. Find the composition and dew point of the products. The reaction equation for stoichiometric mixture is: CH4 + νO2 (O2 + 3.76 N2) C balance:

νCO2 = 1 ,

→ 2 H2O + 1 CO2 + 3.76 νO2 N2

H balance: 2νH2O = 4 , N2 balance:

2 νO2 = νH2O + 2νCO2 = 2 + 2 × 1

O balance:

νO2 = 1.5 × 2 = 3


CH4 + 3 (O2 + 3.76 N2)

→

=>

νN2 = 3.76 νO2

νO2 = 2


2 H2O + 1 CO2 + 11.28 N2 + 1 O2

Add water to the dry air from Eq.12.28 w = 0.622

φPg Ptot - φPg

= 0.622

0.7 × 3.169 = 0.0141 100 - 0.7 × 3.169

So the number of moles to add is from Eq.14.9 x = 7.655 w = 7.655 × 0.0141 = 0.108 and the added number of moles is Products:

νO2 x = 0.324, the products are then

2.324 H2O + 1 CO2 + 11.28 N2 + 1 O2

The water partial pressure becomes Pv = yv Ptot =

2.324 100 = 14.894 kPa 2.324 + 1 + 11.28 + 1

Tdew = 53.8oC



15.121

Pentane is burned with 120% theoretical air in a constant pressure process at 100 kPa. The products are cooled to ambient temperature, 20 °C. How much mass of water is condensed per kilogram of fuel? Repeat the answer, assuming that the air used in the combustion has a relative humidity of 90%. C5H12 + 1.2 × 8 (O2 + 3.76 N2) → 5 CO2 + 6 H2O + 0.96 O2 + 36.1 N2 Products cooled to 20oC, 100 kPa, so for H2O at 20°C: P g = 2.339 kPa nH2O MAX 2.339 yH2O MAX = Pg/P = = ⇒ nH2O MAX = 1.007 < nH2O 100 nH2O MAX + 42.06 Therefore,

nH2O VAP = 1.007,

mH2O LIQ =

4.993 × 18.015 = 1.247 kg/kg fuel 72.151

Pv1 = 0.9 × 2.339 = 2.105 kPa nH2O IN = 0.013375 ×

⇒

w1 = 0.622 ×

2.105 = 0.013375 97.895

28.97 × (9.6 + 36.1) = 0.983 kmol 18.015

nH2O OUT = 0.983 + 6 = 6.983 mH2O LIQ =

nH2O LIQ = 6 - 1.007 = 4.993

5.976 × 18.015

⇒

nH2O LIQ = 6.983 - 1.007 = 5.976 kmol

= 1.492 kg/kg fuel

72.151



15.122

A gas turbine burns methane with 150% theoretical air. Assume the air is 25 oC, 100 kPa and relative humidity of 80%. How large a fraction of the product mixture water comes from the moist inlet air? The reaction equation for stoichiometric mixture is: CH4 + νO2 (O2 + 3.76 N2) C balance: O balance:

→ 2 H2O + 1 CO2 + 3.76 νO2 N2

νCO2 = 1 , H balance: 2νH2O = 4 , 2 νO2 = νH2O + 2νCO2 = 2 + 2 × 1 νO2 = 1.5 × 2 = 3


CH4 + 3 (O2 + 3.76 N2)

→

N2 balance: =>

νN2 = 3.76 νO2

νO2 = 2


2 H2O + 1 CO2 + 11.28 N2 + 1 O2

Add water to the dry air from Eq.12.28 w = 0.622

φPg Ptot - φPg

= 0.622

0.7 × 3.169 = 0.0141 100 - 0.7 × 3.169

So the number of moles to add is from Eq.15.??9 x = 7.655 w = 7.655 × 0.0141 = 0.108 and the added number of moles is Products:

νO2 x = 0.324, the products are then

2.324 H2O + 1 CO2 + 11.28 N2 + 1 O2

Fraction of water is 0.324/2.324 = 0.1394 so 13.9%



15.123

In an engine a mixture of liquid octane and ethanol, mole ratio 9:1, and stoichiometric air are taken in at T0, P0. In the engine the enthalpy of combustion is used so that 30% goes out as work, 30% goes out as heat loss and the rest goes out the exhaust. Find the work and heat transfer per kilogram of fuel mixture and also the exhaust temperature. 0.9 C8H18 + 0.1 C2H5OH + 11.55 O2 + 43.428 N2

→ 8.4 H2O + 7.4 CO2 + 43.428 N2 For 0.9 octane + 0.1 ethanol, convert to mole basis ° ° -° HRP mix= 0.9 HRP C8H18 + 0.1 HRP C2H5OH

= 0.9 (-44 425) ×114.232 + 0.1 (-26 811) × 46.069 = -4 690 797

kJ kmol

^ ^ ^ = 107.414 M = 0.9 M + 0.1 M mix oct alc Energy:

-° -° hin + qin = hex + ωex = hex + ∆hex + ωex

-° -° -° hex - hin = HRP mix

-

-°

⇒ ωex + ∆hex - qin = -HRP mix

-°

kJ

kJ

ωex= -qin = 0.3(-HRP)= 1 407 239 kmol = 13 101 kg Fu -

-

-°

kJ

∆hprod = ∆hex = 0.4(-HRP)= 1 876 319 kmol Fu ∆h-prod = 8.4 ∆h-H2O + 7.4∆h-CO2 + 43.428 ∆h-N2 -

∆hprod 1300 = 8.4× 38 941 + 7.4× 50 148 + 43.428× 31 503 = 2 066 312 -

∆hprod 1200 = 8.4× 34 506 + 7.4× 44 473 + 43.428× 28 109 = 1 839 668 Linear interpolation to get the value of

⇒

-

∆hprod = 1 876 319

satisfied for T = 1216 K



15.124

The gas turbine cycle in problem 12.21 has a q H = 960 kJ/kg air added by combustion. Assume the fuel is methane gas and the q H equals the heating value at To. Find the A/F ratio on a mass basis.

The heating value and the q H are scaled with the fuel mass and total mixture mass respectively so the relation is

qH =

HV HV = = 960 kJ/kg-mix mmix/mfuel A/F + 1

A/F =

HV 50 010 − 1 = 960 − 1 = 51.09 qH

Comment: This is about 300% theoretical air.



15.125

A gas turbine burns methane with 200% theoretical air. The air and fuel comes in through two separate compressors bringing them from 100 kPa, 298 K to 1400 kPa and enters a mixing chamber and combustion chamber. What are the specific compressor work and q H to be used in Brayton cycle calculation? Use constant specific heat to solve. The reaction equation for stoichiometric mixture is: CH4 + νO2 (O2 + 3.76 N2) O balance:

2 νO2 = 2 + 2

=>

νO2 = 2

νO2 = 2 × 2 = 4


CH4 + 4 (O2 + 3.76 N2)

→ 2 H2O + 1 CO2 + c N2 so now more O2 and N2

→ 2 H2O + 1 CO2 + 15.04 N2 + 2 O2

m = 16.043 + 4 × 4.76 × 28.97

For 1 kmol fuel:

= 16.043 + 551.59 = 567.63 kg reactants The temperatures after compression are: T2 air = T1 (P2/P1)

(k-1)/k

T2 fuel = T1 (P2/P1)

= 298.15 × 14

(k-1)/k

0.2857

= 298.15 × 14

= 633.7 K

0.230177

(k = 1.4)

= 547.3 K (k = 1.299)

wc air = Cp (T2 – T1) = 1.004(633.7 – 298.15) = 336.89 kJ/kg air wc fuel = Cp (T2 – T1) = 2.254(547.3 – 298.15) = 561.68 kJ/kg fuel wc total =

mfuel mair 16.04 551.59 wc fuel + wc air = 561.58 + 336.89 m m 567.63 567.63

= 343.24 kJ/kg mix °

q = -HRP = HV but scaled to be per kg reactants q=

50 010 × 16.043 = 1413.4 kJ/kg 567.63

(=

HV ) A/F + 1



15.126

Find the equivalent heat transfer q H to be used in a cycle calculation for constant pressure combustion when the fuel is (a) methane and (b) gaseous octane in both cases use water vapor in the products and a stoichiometric mixture. Methane combustion becomes CH4 +

νO2 (O2 + 3.76 N2 ) → 1 CO2 + 2 H2O + 7.52 N2 2νO2 = 2 + 2 = 4 ⇒ νO2 = 2, νN2 = 7.52

Oxygen balance:

2 × 31.999 + 7.52 × 28.013 = 17.12 = mair/mfuel 16.043 HV HV 50 010 q= = = = 2760 kJ/kg mmix/mfuel A/F + 1 17.12 + 1 A/F =

Octane combustion, see reaction and A/F in Example 15.1, HV from Table 15.3 q=

HV HV 44 788 = = = 2799 kJ/kg mmix/mfuel A/F + 1 15 + 1



15.127

Consider the steady combustion of propane at 25 °C with air at 400 K. The products exit the combustion chamber at 1200 K. It may be assumed that the combustion efficiency is 90%, and that 95% of the carbon in the propane burns to form carbon dioxide; the remaining 5% forms carbon monoxide. Determine the ideal fuel–air ratio and the heat transfer from the combustion chamber. Ideal combustion process, assumed adiabatic, excess air to keep 1200 K out. C3H8 + 5x O2 + 18.8x N2 → 3 CO2 + 4 H2O + 5(x - 1) O2 + 18.8x N2 HR = -103900 +5x(0 + 3027) + 18.8x(0 + 2971) = -103900 + 70990x HP = 3(-393522 + 44473) + 4(-241826 + 34506) + 5(x - 1)(0 + 29761) + 18.8x(0 + 28109) = -2025232 + 677254x 1st law: HP - HR = 0

Solving, x = 3.169

FAIDEAL = 1/(23.8 × 3.169) = 0.01326 b)

FAACTUAL = 0.01326/0.90 = 0.01473 C3H8 + 14.26 O2 + 53.62 N2

→ 2.85 CO2 + 0.15 CO + 4 H2O + 9.335 O2 + 53.62 N2 HR = -103900 + 14.26(0 + 3027) + 53.62(0 + 2971) = +98 570 kJ HP = 2.85(-393522 + 44473) + 0.15(-110527 + 28427) + 4(-241826 + 34506) + 9.335(0 + 29761) + 53.62(0 + 28109) = -51 361 kJ QCV = HP - HR = -149 931 kJ



15.128

A gasoline engine is converted to run on propane. Assume the propane enters the engine at 25°C, at the rate 40 kg/h. Only 90% theoretical air enters at 25 °C such that 90% of the C burns to form CO 2, and 10% of the C burns to form CO. The combustion products also include H2O, H2 and N2, exit the exhaust at 1000 K. Heat loss from the engine (primarily to the cooling water) is 120 kW. What is the power output of the engine? What is the thermal efficiency? . Propane: T1 = 25oC, m = 40 kg/hr, M = 44.094 kg/kmol Air: T2 = 25oC, 90% theoretical Air produces 90% CO2, 10% CO Products: T3 = 1000 K, CO2, CO, H2O, H2, N2 C3H8 + 4.5O2 + 16.92N2  2.7 CO2 + 0.3CO + 3.3H2O + 0.7H2 + 16.92N2 . . nC3H8 = m/(M×3600) = 0.000252 kmol/s . . . 1st Law: Q + HR = HP + W ; Q = -120 kW

−o

HR = nC3H8 h f = -103 900 kJ Products:

−o

−

−o

−

CO2

-

nCO2(h f + ∆h ) = 2.7(-393522 + 33397) = -972337.5 kJ

CO

-

nCO(hf + ∆h) = 0.3(-110527 + 21686) = -26652 kJ

H2O

-

nH2O(− hof + ∆− h) = 3.3(-241826 + 26000) = -712226 kJ

H2

-

nH2(h f + ∆h ) = 0.7(0 + 20663) = 14464.1 kJ

N2

-

nN2(h f + ∆h ) = 16.92(0 + 21463) = 363154 kJ

−o

−o

−

−o

−

−

HP = ni (h f + ∆h)i = -1 333 598 kJ . . . W = Q + n(HR - HP) = 189.9 kW

∑

C3H8: Table 14.3 HRPo = -50343 kJ/kg . . HHV = nC3H8 M(-HRPo) = 559.4 kW . . ηth = W/HHV = 0.339



15.129

A small air-cooled gasoline engine is tested, and the output is found to be 1.0 kW. The temperature of the products is measured to 600 K. The products are analyzed on a dry volumetric basis, with the result: 11.4% CO2, 2.9% CO, 1.6% O2 and 84.1% N2. The fuel may be considered to be liquid octane. The fuel and air enter the engine at 25°C, and the flow rate of fuel to the engine is 1.5 × 10-4 kg/s. Determine the rate of heat transfer from the engine and its thermal efficiency. a C8H18 + b O2 + 3.76b N2

→ 11.4 CO2 + 2.9 CO + c H2O + 1.6 O2 + 84.1 N2 b=

84.1 = 22.37, 3.76

a=

1 (11.4 + 2.9) = 1.788 8

c = 9a = 16.088 C8H18 + 12.5 O2 + 47.1 N2

→ 6.38 CO2 + 1.62 CO + 9 H2O + 0.89 O2 + 47.1 N2 -0 HR = hf C8H18 = -250 105 kJ/kmol HP = 6.38(-393 522 + 15 788) + 1.62(-110527 + 10 781) + 9(-241 826 + 12 700) + 0.89(0 + 11187) + 47.1(0 + 10712) = -4 119 174 kJ/kmol HP - HR = -4 119 174 - (-250 105) = -3 869 069 kJ/kmol . . HP - HR = (0.00015/114.23)(-3 869 069) = -5.081 kW . QCV = -5.081 + 1.0 = -4.081 kW Fuel heating value from table 15.3 . . QH = m HV = 0.00015 (47 893) = 7.184 kW . . ηTH = WNET/QH = 1.0/7.184 = 0.139



15.130

A gasoline engine uses liquid octane and air, both supplied at P , T , in a 0 0 stoichiometric ratio. The products (complete combustion) flow out of the exhaust valve at 1100 K. Assume that the heat loss carried away by the cooling water, at 100°C, is equal to the work output. Find the efficiency of the engine expressed as (work/lower heating value) and the second law efficiency. C8H18 + νO (O2 + 3.76 N2) → 8 CO2 + 9 H2O + 47 N2 2

2 νO = 16 + 9 ⇒ νO = 12.5 2

2

LHV = 44425

kJ kg fuel

⇒

kJ

LHV = 5.07476×106 kmol fuel

∆HP 1100 = 8 × 38885 + 9 × 30190 + 47 × 24760 = 1746510 C.V. Total engine Hin = Hex + W + Qloss = Hex + 2 W °

⇒ 2 W = Hin - Hex= HR - Hν= -HRP + ∆HR- ∆HP 1100 = 5.07476×106 + 0 - 1746510 = 3328250 kJ

W = 1.664×106 kmol fuel

ηth =

W LHV

1.664×106

=

6

= 0.328

5.07476×10

Find entropies in and out: inlet: -° SFu = 360.575 1 SO° = 205.148 - 8.3145 ln = 218.12 4.76 2 3.76 SN° = 191.609 - 8.3145 ln = 193.57 4.76 2 -° Sin = 360.575 + 12.5 × 218.12 + 47 × 193.57 = 12185 8 - ° exit: SCO = 275.528 - 8.3145 ln = 292.82 64 2 S-H°O = 236.732 - 8.3145 ln 9 = 253.04 64 2 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Borgnakke and Sonntag 47 SN° = 231.314 - 8.1345 ln = 233.88 64 2 -° Sex = 8 × 292.82 + 9 × 253.04 + 47 × 233.88 = 15612 Assume the same Qloss out to 100°C reservoir in the reversible case and rev

compute Q0 : rev Sin + Q0 /T0 = Sex + Qloss/Tres

Qrev 0

= T0(S-ex - S-in) + Qloss T0/Tres = 298.15(15612 - 12185) + 1.664 ×106 × 298.15/373.15 kJ

= 2.351×106 kmol fuel rev

Hin + Q0 = Hex + Wrev + Qloss

⇒ Wrev

rev

rev

kJ

= Hin - Hex - Qloss + Q0 = Wac + Q0 = 4.015×106 kmol fuel

ηII = Wac/ Wrev =1.664×106/4.015×106 = 0.414



Review Problems



15.131

Repeat Problem 15.25 for a certain Utah coal that contains, according to the coal analysis, 68.2% C, 4.8% H, 15.7% O on a mass basis. The exiting product gas contains 30.9% CO, 26.7% H 2, 15.9% CO2 and 25.7% H2O on a mole basis. Convert the mass concentrations to number of kmol per 100 kg coal: C : 68.2/12.01 = 5.679

H2: 4.8/2.016 = 2.381

O2: 15.7/32.00 = 0.491 Now the combustion equation reads x(5.679 C + 2.381 H2 + 0.491 O2) + y H2O + z O2 in 30.9 CO + 26.7 H2 + 15.9 CO2 + 25.7 H2O out in 100 kmol of mix out Now we can do the atom balance to find (x, y, z)

→

C:

5.679x = 30.9 + 15.9

H2:

2.381 × 8.241 + y = 26.7 + 25.7

O2:

32.778 30.9 25.7 0.491 × 8.241 + +z= + 15.9 + 2 2 2

→

x = 8.241

→

y = 32.778

z = 23.765

Therefore, for 100 kmol of mixture out, require: 824.1 kg of coal 32.778 kmol of steam 23.765 kmol of oxygen



15.132

Many coals from the western United States have a high moisture content. Consider the following sample of Wyoming coal, for which the ultimate analysis on an as-received basis is, by mass: Component Moisture H C S N O Ash % mass 28.9 3.5 48.6 0.5 0.7 12.0 5.8 This coal is burned in the steam generator of a large power plant with 150% theoretical air. Determine the air–fuel ratio on a mass basis. Converting from mass analysis: Substance

S

H2

C

O2

N2

c/M =

0.5/32

3.5/2

4.86/12

12/32

0.7/28

kmol / 100 kg coal

0.0156

1.75

4.05

0.375

0.025

Product

SO2

H2O

CO2

oxygen required

0.0156

0.875

4.05

--

--

Combustion requires then oxygen as: 0.0156 + 0.875 + 4.05 = 4.9406 The coal does include 0.375 O2 so only 4.5656 O2 from air/100 kg coal AF = 1.5 × (4.5656 + 4.5656 × 3.76) × 28.97/100 = 9.444 kg air/kg coal



15.133

A fuel, CxHy, is burned with dry air and the product composition is measured on a dry mole basis to be: 9.6% CO2, 7.3% O2 and 83.1% N2. Find the fuel composition (x/y) and the percent theoretical air used.

νFuCxHy + νO2O2 + 3.76νO2N2 → 9.6 CO2 + 7.3 O2 + 83.1 N2 + νH2OH2O N2 balance:

3.76νO = 83.1 ⇒ νO = 22.101

O2 balance:

νO2 = 9.6 + 7.3 + 2 νH2O ⇒ νH2O = 10.402

H balance:

νFu y = 2 νH2O = 20.804

C balance:

νFu x = 9.6

2

1

2

Fuel composition ratio = x/y = 9.6/20.804 = 0.461 Theoretical air =

νO

2AC

νO

2stoich

100 =

22.101 1

9.6 + 4 × 29.804

100 = 149.3%



15.134

In an engine liquid octane and ethanol, mole ration 9:1, and stoichiometric air are taken in at 298 K, 100 kPa. After complete combustion, the products run out of the exhaust system where they are cooled to 10 oC. Find the dew point of the products and the mass of water condensed per kilogram of fuel mixture. Reaction equation with 0.9 octane and 0.1 ethanol is 0.9 C8H18 + 0.1 C2H5OH + 11.55 O2 + 43.428 N2

→ 8.4 H2O + 7.4 CO2 + 43.428 N2 yH

2O

PH

2O

=

8.4 = 0.1418 8.4 + 7.4 + 43.428

= yH

P

2O tot

10 °C ⇒ PH

2O

⇒

= 14.3 kPa

= 1.2276

⇒

yH

2O

Tdew= 52.9 °C

= 0.012276 =

x x + 7.4 + 43.428

kmol

⇒ x = 0.6317 ⇒ ∆νH2O= -7.77 kmol Fu mix Mmix = mH

∑yiMi = 0.9 × 114.23 + 0.1 × 46.069 = 107.414

= O cond

2

-∆νH

2O

× 18.015

107.414

kg

= 1.303 kg Fu mix



15.135

In a test of rocket propellant performance, liquid hydrazine (N2H4) at 100 kPa, 25°C, and oxygen gas at 100 kPa, 25 °C, are fed to a combustion chamber in the ratio of 0.5 kg O2/kg N2H4. The heat transfer from the chamber to the surroundings is estimated to be 100 kJ/kg N 2H4. Determine the temperature of the products exiting the chamber. Assume that only H 2O, H2, and N2 are present. The enthalpy of formation of liquid hydrazine is +50 417 kJ/kmol. 1

Liq. N2H4: 100 kPa, 25oC

3

Comb. Chamber

o

Gas O2: 100 kPa, 25 C

Products

2

. . . . mO2/mN2H4 = 0.5 = 32nO2/32nN2H4

Energy Eq.:

. . Q/mN2H4 = -100 kJ/kg

and

QCV = HP - HR = -100 × 32.045 = -3205 kJ/kmol fuel 1

1 N2H4 + 2 O2 → H2O + H2 + N2

Combustion eq.: 1

HR = 1(50417) + 2(0) = 50417 kJ HP = -241 826 + ∆hH2O + ∆hH2 + ∆hN2 Energy Eq. now reads o HP = HR + QCV = HP + -

-

-

∆HP o

∆HP = ∆hH2O + ∆hH2 + ∆hN2 = -HP +

HR + QCV

= 241 826 + 50 417 - 3205 = 289 038 kJ/kmol fuel Table A.9 : Guess T and read for water, hydrogen and nitrogen 2800 K:

∆HP = 115 463 + 81 355 + 85 323 = 282 141

too low

3000 K:

∆HP = 126 548 + 88 725 + 92 715 = 307 988

too high

Interpolate to get

TP = 2854 K



15.136

Find the lower heating value for the fuel blend in problem 15.134 with a scaling as in Table 15.3 Reaction equation with 0.9 octane and 0.1 ethanol is 0.9 C8H18 + 0.1 C2H5OH + 11.55 O2 + 43.428 N2

→ 8.4 H2O + 7.4 CO2 + 43.428 N2 °

We can solve several ways, use Table A.10 to find H RP and do it on a mole basis and then divide with M for mixture or we can convert the mole fractions to mass fractions and use Table 15.3 values. Let us use Table 15.3. Mmix =

∑yiMi = 0.9 × 114.23 + 0.1 × 46.069 = 107.414

0.1 × 46.069 = 0.042889, coctane = 0.957111 107.414 LHV = cethanol LHVethanol + coctane LHVoctane = 0.042889 (26 811) + 0.957111 (44 425) kJ/kg = 43 670 kJ/kg cethanol = yM/Mmix =



15.137

E85 is a liquid mixture of 85% ethanol and 15% gasoline (assume octane) by mass. Find the lower heating value for this blend. The heating value of the blend becomes LHV = 0.85 LHVethanol + 0.15 LHVoctane = 0.85 (26 811) + 0.15 (44 425) kJ/kg = 29 453 kJ/kg As this is lower than gasoline a larger amount of fuel is being used for the same energy release in the engine. If the mixture was given on a mole basis we would have used LHV on a mole basis instead (multiplying Table 15.3 entries by M).



15.138

Determine the higher heating value of the sample Wyoming coal as specified in Problem 15.132. The higher heating value is for liquid water in the products. We need the composition of the products. Converting from mass analysis:

Substance c/M =

S 0.5/32

H2 3.5/2

C 4.86/12

O2 12/32

N2 0.7/28

kmol / 100 kg coal

0.0156

1.75

4.05

0.375

0.025

Product

SO2

H2O

CO2

So the combustion equation becomes (for 100 kg coal) Fuel + Air

→

1.75 H2O + 4.05 CO2 + 0.0156 SO2

The formation enthalpies are from Table A.10. Therefore, o o hRP0 = HP - HR = 4.05(-393 522) + 1.75(-285 830) + 0.0156(-296 842) = -2 098 597 kJ/100 kg coal So that HHV = +20 986 kJ/kg coal



15.139

Ethene, C2H4, and propane, C3H8, in a 1:1 mole ratio as gases are burned with 120% theoretical air in a gas turbine. Fuel is added at 25 °C, 1 MPa and the air comes from the atmosphere, 25°C, 100 kPa through a compressor to 1 MPa and mixed with the fuel. The turbine work is such that the exit temperature is 800 K with an exit pressure of 100 kPa. Find the mixture temperature before combustion, and also the work, assuming an adiabatic turbine.

φ = 1:

C2H4 + C3H8 + 8 O2 + 30.08 N2 → 5 CO2 + 6 H2O + 30.08 N2

φ = 1.2:

C2H4 + C3H8 + 9.6 O2 + 36.096 N2

→ 5 CO2 + 6 H2O + 1.6 O2 + 36.096 N2 45.696 kmol air per 2 kmol fuel C.V. Compressor (air flow) Energy Eq.: Pr = Pr 2

w c = h2 - h1

Entropy Eq.:

× P2/P1 = 13.573

⇒

1

s2 = s1

⇒

T2 air = 570.8 K

wc = 576.44 - 298.34 = 278.1 kJ/kg = 8056.6 kJ/kmol air C.V. Mixing Chanber (no change in composition) . . . nairhair in + nFu1h1 in + nFu2h2 in = (SAME)exit (CP F1 + CP F2)(Texit - T0) = 45.696 CP air(T2 air - Texit) C = 43.43, P F1

C2H4:

C3H8: C = 74.06, P F2

C = 29.07 P air

45.696CP airT2 + (CP F1 + CP F2)T0 Texit = = 548.7 K CP F1 + CP F2 + 45.696 CP air

(

)

Dew Point Products: yH PH

2O

= yH

P

2O tot

6

2O

= 5 + 6 + 1.6 + 36.096 = 0.1232

= 123.2 kPa

⇒

Tdew = 105.5°C

C.V. Turb. + combustor + mixer + compressor (no Q) wnet

= Hin - Hout = HR - HP 800

(800°K out so no liquid H 2O)

-° -° = hf C2H4 + hf C3H8 - 5 hCO - 6 hH 2

= 2 576

2O

- 1.6 hO - 36.096 hN 2

2

kJ 541 2 kmol Fu

wT = wnet + wcomp = 2 944 695

kJ 2 kmol Fu



15.140

A study is to be made using liquid ammonia as the fuel in a gas-turbine engine. Consider the compression and combustion processes of this engine. a. Air enters the compressor at 100 kPa, 25 °C, and is compressed to 1600 kPa, where the isentropic compressor efficiency is 87%. Determine the exit temperature and the work input per kilomole. b. Two kilomoles of liquid ammonia at 25 °C and x times theoretical air from the compressor enter the combustion chamber. What is x if the adiabatic flame temperature is to be fixed at 1600 K? Air P1 = 100 kPa

1

2

P2 = 1600 kPa

COMP.

ηS COMP = 0.87

o

T1 = 25 C

-W

a) ideal compressor process (adiabatic reversible): P2 k-1 1600 0.286 k s2S = s1 ⇒ T2S = T1( ) = 298.2( ) = 659 K P1 100 -wS = CP0(T2S - T1) = 1.004(659 - 298.2) = 362.2 Real process: -w = -w /η = 362.2/0.87 = 416.3 kJ/kg S

S

T2 = T1 - w/CP0 = 298.2 + 416.3/1.004 = 713 K Also -w = 416.3 × 28.97 = 12060 kJ/kmol b) 2 liq NH3, 25 oC Air 100 kPa, 25 oC

COMP -W

COMB. CHAMBER

Prod. PP = 1600 kPa TP = 1600 K

-Q = 0

2 NH3 + 1.5x O2 + 5.64x N2 → 3 H2O + 1.5(x - 1) O2 + (5.64x + 1) N2 Using Tables 15.3, A.10 and A.2, hNH3 = -45 720 + 17.031(298.36 - 1530.04) = -66 697 kJ/kmol HR = 2(-66 697) + 0 = -133 394 kJ -W = 12 060 × 7.14x = 86 108 x kJ Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.


HP = 3(-241 826 + 52 907) + 1.5(x - 1)(0 + 44267) + (5.64x + 1)(0 + 41904) = 302 739x – 591 254 Energy Eq.:

HR = HP + W -133 394 = 302 739 x – 591 254 – 86 108 x

⇒

x = 2.11



15.141

Consider the gas mixture fed to the combustors in the integrated gasification combined cycle power plant, as described in Problem 15.30. If the adiabatic flame temperature should be limited to 1500 K, what percent theoretical air should be used in the combustors? Product

CH4

H2

CO

CO2

N2

H 2O

H2S

NH3

% vol.

0.3

29.6

41.0

10.0

0.8

17.0

1.1

0.2

Mixture may be saturated with water so the gases are ( H2S and NH3 out) CH4 H2 CO CO2 N2 n 0.3

29.6

41.0

10.0

0.8

81.7

yV MAX = 7.384/3000 = nV/(nV + 81.7) Solving, nV = 0.2 kmol, the rest is condensed

{0.3 CH4 + 29.6 H2 + 41.0 CO + 10.0 CO2 + 0.8 N2 + 0.2 H2O + 35.9x O2 + 3.76 × 35.9x N2}→ 51.3 CO2 + 30.4 H2O + 35.9(x - 1) O2 + (135.0x + 0.8) N2 For the fuel gas mixture, nCP0 MIX = 0.3 × 16.04 × 2.2537 + 29.6 × 2.016 × 14.2091 + 41.0 × 28.01 × 1.0413 + 10.0 × 44.01 × 0.8418 + 0.8 × 28.013 × 1.0416 + 0.2 × 18.015 × 1.8723 = 2455.157 -0 nhf MIX = 0.3(-74873) + 29.6(0) + 41.0(-110527) + 10.0(-393522) + 0.8(0) + 0.2(-241826) = -8537654 kJ At 40°C, for the fuel mixture: HMIX = -8 537 654 + 2455.157(40 - 25) = -8 500 827 kJ Assume air enters at 25°C: hAIR = 0 Products at 1500 K: HP = 51.3(-393522 + 61705) + 30.4(-241826 + 48149) + 35.9(x - 1)(0 + 40600) + (135x + 0.8)(0 + 38405) = -24 336 806 + 6 642 215x Energy Eq.:

HP = HR = HMIX



x=

+24336809 - 8500827 = 2.384 6642215

or

238 % theo. air



15.142

Carbon monoxide, CO, is burned with 150% theoretical air and both gases are supplied at 150 kPa and 600 K. Find the reference enthalpy of reaction and the adiabatic flame temperature. CO +

νO

(O2 + 3.76 N2) →

2

O balance:

1 + 2 νO = 1

CO2 + νN N2 2

⇒ νO2 = 0.5 ⇒ νO2 actual = 0.75

2

Now the actual reaction equation has excess oxygen as CO + 0.75 (O2 + 3.76 N2) → CO2 + 2.82 N2 + 0.25 O2 From the definition of enthalpy of combustion, Eq.14.14 or 14.15 ° ° ° -o -o HRP = HP - HR = hf CO2 + 0 - hf CO

= -393 522 – (-110 527) = -282 995 kJ/kmol CO = - 10 103 kJ/kg CO

(as for Table 14.3)

°

°

HR = HP = HP + ∆HP = HR + ∆HR

Actual energy Eq.: °

°

°

-

-

-

∆HP = HR + ∆HR - HP = -HRP + ∆hCO + 0.75 ∆hO2 + 2.82 ∆hN2 = 282 995 + 8942 + 0.75

× 9245 + 2.82 × 8894

= 323 952 kJ/kmol The left hand side is ∆HP = ∆hCO + 0.25 ∆hO + 2.82 ∆hN 2

2

2

∆HP 2600= 128 074 + 0.25 × 82 225 + 2.82 × 77 963 = 368 486 ∆HP 2400= 115 779 + 0.25 × 74 453 + 2.82 × 70 640 = 333 597 ∆HP 2200= 103 562 + 0.25 × 66 770 + 2.82 × 63 362 = 298 935 Now we can do a linear interpolation for the adiabatic flame temperature T = 2200 + 200

323 952 - 298 935 = 2344 K 333 597 - 298 935



15.143

A rigid container is charged with butene, C 4H8, and air in a stoichiometric ratio at P0, T0. The charge burns in a short time with no heat transfer to state 2. The products then cool with time to 1200 K, state 3. Find the final pressure, P3, the total heat transfer, 1Q3, and the temperature immediately after combustion, T2. The reaction equation is, having used C and H atom balances: C4H8 + νO (O2 + 3.76 N2) → 4 CO2 + 4 H2O + 3.76 νO N2 2

2

Counting now the oxygen atoms we require

νO2= 6.

C.V. analysis gives: U3 - U1 = Q - W = Q = H 3 - H1 - P2V2 + P1V1 = H3 - H1 - R(n2T3 - n1T1) ° ° ° ^ H° + ∆H H3 - H1 = HP 1200 - HR = HP - HR + ∆HP = M RP P

= 56.108 × (-45 316) + 950055 = -1 592 535 kJ/kmol ^ = 56.108 and Where M n1= 1 + 6 × 4.76 = 29.56, n2 = 4 + 4 + 6 × 3.76 = 30.56,

Τable A.9 at 1200 K:

∆hCO

2

= 44473,

∆hH

2O

=34506,

∆hN

=28109.

2

Now solving for the heat transfer: Q = -1592535 - 8.3145(30.56

kJ

× 1200 - 29.56 × 298.15) = -1824164 kmol fuel

To get the pressure, assume ideal gases: n2RT3 n2T3 P3 = = P1 = 421.6 kPa V2 n1T1 Before heat transfer takes place we have constant U so: U2 - U1 = 0 = H2 - H1 - n2RT2 + n1RT1 °

Now split the enthalpy H2= HP + ∆HP(T2) and arrange things with the unknowns on LHS and knowns on RHS: ° ∆HP- n2RT2 = HR - HP - n1RT1= 2 542 590 - 73278 = 2 469 312 Trial and error leads to:

× 8.31451 × 3000 = 2 446 980 LHS (3200 K) = 3 471 331 - 30.56 × 8.31451 × 3200 = 2 658 238 LHS (3000 K) = 3 209 254 - 30.56 linear interpolation

T = 3021 K



15.144

Natural gas (approximate it as methane) at a ratio of 0.3 kg/s is burned with 250% theoretical air in a combustor at 1 MPa where the reactants are supplied at T0. Steam at 1 MPa, 450°C at a rate of 2.5 kg/s is added to the products before they enter an adiabatic turbine with an exhaust pressure of 150 kPa. Determine the turbine inlet temperature and the turbine work assuming the turbine is reversible. CH4 + νO (O2 + 3.76 N2) → CO2 + 2 H2O + 7.52 N2 2

2 νO = 2 + 2

⇒ νO

2

2

=2

Actual νO = 2 × 2.5 = 5

=>

2

CH4 + 5 O2 + 18.8 N2 → CO2 + 2 H2O + 3 O2 + 18.8 N2 C.V. combustor and mixing chamber HR + nH OhH O in = HP ex 2

2

. . nH2O mH2OMFu 2.5 × 16.043 kmol steam . nH O = = . = = 7.421 kmol fuel nFu mFuMH2O 0.3 × 18.015 2 Energy equation becomes nH O(hex - hin)H O + 2

-

-

-

-

(∆hCO2 + 2∆hH2O + 3∆hO2 + 18.8∆hN2)ex

2

°

h

(

(

= -HRP = 50 010 × 16.043 = 802 310 = ∆h - 15072.5, so then:

-h ex

in H2O

H2O ex

) ∆hCO + 9.421∆hH

2O

2

+ 3∆hO + 18.8∆hN 2

kJ

)

2 ex

= 914 163 kmol fuel

Trial and error on Tex Tex = 1000 K

⇒

LHS = 749 956 ;

Tex = 1200 K

⇒

LHS = 987 286

Tex = 1100 K

⇒

⇒

LHS = 867429

Tex ≅ 1139 K = Tin turbine

If air then Tex turbine ≈ 700 K and Tavg ≈ 920 K. Find CP mix between 900 and 1000 K. From Table A.9: CP mix =

+ 3(34.62) + 18.8(32.4) ∑ niC-Pi/∑ ni = 53.67 + 9.421(40.63)32.221

= 35.673 kJ/kmol K =C - R = 27.3587 kJ/kmol, k

C V mix

P mix

Tex turbine = 1139 (150 / 1000)

0.2331

= 1.304 mix

= 732 K



∆H732 = 19370.6 + 9.421(15410) + 3(13567) + 18.8(12932) = 448 371 kJ/kmol kJ

wT = Hin - Hex = ∆Hin - ∆Hex = 914 163 - 448 371 = 465 792 kmol fuel . . • ^ = (0.3 × 465 792)16.043 = 8710 kW WT = nFuwT = mFuwT/M Fu



15.145

The turbine in Problem 15.139 is adiabatic. Is it reversible, irreversible, or impossible? Inlet to the turbine is the exit from the mixing of air and fuel at 1 MPa. From solution to 15.139, we have: CP C H = 43.43, CP C H = 74.06, Tturbine,in = 548.7 K 2 2

3 8

C2H4 + C3H8 + 9.6 O2 + 36.096 N2 → 5 CO2 +6 H2O + 1.6 O2 + 36.096 N2 Sex - Sin = ∫ dQ/T + Sgen = Sgen ≥ φ -° SFu = Si + CP Fu ln(T/T0) s-°i - yi P -Rln P0

Inlet: 1 MPa, 548.7 K

Si

ni

yi

C2H4

1

0.02097

245.82

12.989

258.809

C3H8

1

0.02097

315.09

12.989

328.079

O2

9.6

0.2013

223.497

-5.816

217.681

N2

36.096

0.7568

209.388

-16.828

192.56

S

= 258.809 + 328.079 + 9.6

× 217.681 + 36.096 × 192.56 = 9627.3

in

ni

yi

s-°i

- yi P -Rln P0

S-i

CO2

5

0.1027

257.496

18.925

276.421

H2O

6

0.1232

223.826

17.409

241.235

O2

1.6

0.0329

235.92

28.399

264.319

N2

36.096

0.7413

221.016

2.489

223.505

Sex

= 5 × 276.421 + 6 × 241.235 1.6 × 264.319 kJ

+ 36.096 × 223.505 = 11320 2kmol Fu K Sgen

kJ

= Sex - Sin = 1693 2kmol Fu K> 0

Possible, but one should check the state after combustion to account for generation by combustion alone and then the turbine expansion separately. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.


15.146

Consider the combustion process described in Problem 15.107. a. Calculate the absolute entropy of the fuel mixture before it is throttled into the combustion chamber. b. Calculate the irreversibility for the overall process. From solution to 15.107, fuel mixture 0.8 C2H6 + 0.2 CH4 at 65°C, 10 MPa CP0 FUEL = 49.718 kJ/kmol K. Using Kay’s rule: Tr1 = 1.198, Pr1 = 2.073 and x = 410.4 % theoretical air or 13.13 O2 + 49.36 N2 in at 600 K, 100 kPa and 1.8CO2 + 2.8H2O + 9.93O2 + 49.36N2

out at 100 kPa, 1200 K

* s-0 FUEL = 0.2(186.251) + 0.8(229.597) - 8.3145(0.2 ln 0.2 + 0.8 ln 0.8) = 225.088

338.2

*

10

∆sTP = 49.718 ln 298.2 - 8.3145 ln 0.1 = -32.031 * (s- -s-)FUEL = 1.37 × 8.3145 = 11.391

From Fig. D.3:

s-FUEL = 225.088 - 32.031 - 11.391 = 181.66 kJ/kmol K Air at 600 K, 100 kPa

-Rln(yiP/P0) +12.976

Si 239.426

+1.96

214.137

O2

13.13

0.21

N2

49.36

0.79

212.177

SAIR =

-

yi

-° si 226.45

ni

∑ niS-i = 13713.47 kJ/K

SR = 181.66 + 13713.47 = 13895.1 kJ/K Products at 1200 K, 100 kPa PROD ni yi

-Rln(yiP/P0)

S-i

CO2

1.8

0.0282

o s-i 279.390

+29.669

309.059

H2O

2.8

0.0438

240.485

+26.008

266.493

O2

9.93

0.1554

250.011

+15.479

265.490

N2

49.36

0.7726

234.227

+2.145

236.372

SP =

∑ niSi = 15606.1 kJ/K


Borgnakke and Sonntag I = T0(SP - SR) - QCV = 298.15(15 606.1 – 13 895.1) + 0 = 510 132 kJ



15.147

Consider one cylinder of a spark-ignition, internal-combustion engine. Before the compression stroke, the cylinder is filled with a mixture of air and methane. Assume that 110% theoretical air has been used, that the state before compression is 100 kPa, 25°C. The compression ratio of the engine is 9 to 1. a. Determine the pressure and temperature after compression, assuming a reversible adiabatic process. b. Assume that complete combustion takes place while the piston is at top dead center (at minimum volume) in an adiabatic process. Determine the temperature and pressure after combustion, and the increase in entropy during the combustion process. c. What is the irreversibility for this process? 1 CH4 + 1.1 × 2 O2 + 3.76 × 2.2 N2 → 1 CO2 + 2 H2O + 0.2 O2 + 2 H2O P1 = 100 kPa, T1 = 298.2 K,

V2/V1 = 1/8, Rev. Ad. s2 = s1

Assume T2 ~ 650 K → TAVE ~ 475 K Table A.6: CP0 CH4 = 44.887, CP0 O2 = 30.890, CP0 N2 = 29.415 CP0 MIX = (1 × 44.887 + 2.2 × 30.890 + 8.27 × 29.415)/11.47 = 31.047 CV0 MIX = CP0 - R = 22.732, k = CP0/CV0 = 1.366 a) T2 = T1(V1/V2)k-1 = 298.2 (9)0.366 = 666.4 K (avg OK) P2 = P1(V1/V2)k = 100 (9)1.366 = 2011 kPa b) comb. 2-3 const. vol., Q = 0 Q = 0 = (H3 - H2) - R(n3T3 - n2T2) 2 3 -0 H2 = 1 hf CH4 + n2 CP0 MIX (T2 - T1) H2 = -74873 + 11.47 × 31.047(666.4 -298.2) = +56246 kJ -* -* -* -* H3 = 1(-393522 + ∆hCO2) + 2(-241826 + ∆hH2O) + 0.2 ∆hO2 + 8.27 ∆hN2 Substituting, -* -* -* -* 1 ∆hCO2 + 2 ∆hH2O + 0.2 ∆hO2 + 8.27 ∆hN2 - 95.366 T3 - 869868 = 0 Trial & error: T3 = 2907 K 1 × 147072 + 2 × 121377 + 0.2 × 94315 + 8.27 × 89274 - 95.366 × 2907 - 869868 ≈ 0 OK Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.


P3 = P2

n3T3 T3 2907 = P2 = 2011 × = 8772 kPa n2T2 T2 666.4

c) state 1 REAC

ni

yi

s-°i

- R ln(yiP/P0)

S-i

CH4

1

0.0872

186.251

+20.283

206.534

O2

2.2

0.1918

205.148

+13.730

218.878

8.27 0.7210 191.609 11.47 S2 = S1 = niSi = 2295.17 kJ/K

+2.720

194.329

N2

∑

state 3 PROD

ni

yi

s-°i

- R ln(yiP/P0

S-i

CO2

1

0.0872

332.213

-16.916

315.297

H2O

2

0.1744

284.753

-22.680

262.073

O2

0.2

0.0174

283.213

-3.516

279.697

8.27

0.7210

265.726

-34.480

231.246

N2 S = 3

∑

11.47 n S = 2807.79 kJ/K i i

S3 - S2 = 2807.79 – 2295.17 = 512.62 kJ/K I = T0(S3 - S2) = 298.2 × 512.62 = 152 860 kJ



15.148

Liquid acetylene, C2H2, is stored in a high-pressure storage tank at ambient temperature, 25°C. The liquid is fed to an insulated combustor/steam boiler at the steady rate of 1 kg/s, along with 140% theoretical oxygen, O 2, which enters at 500 K, as shown in Fig. P15.148. The combustion products exit the unit at 500 kPa, 350 K. Liquid water enters the boiler at 10 °C, at the rate of 15 kg/s, and superheated steam exits at 200 kPa. a.Calculate the absolute entropy, per kmol, of liquid acetylene at the storage tank state. b. Determine theamount phase(s)ofofeach, the combustion exiting the combustor boiler unit, and the if more thanproducts one. c. Determine the temperature of the steam at the boiler exit. -° a) C2H2: SIG 25°C = 200.958 TR = 298.2/308.3 = 0.967 => From Fig. D.1:

PR = 0.82

1

P1 = 0.82 × 6.14 = 5.03 MPa, Sliq T

1 P1

= S°T

0 P0

1

(S-* - S-)1 = 3.33R- = 27.687

- + ∆T - R ln(P1/P°) + (S - S*)P

1 T1

kJ

= 140.695 kmol

K

b) 1 C2H2 + 1.4 × 2.5 O2 → 2 CO2 + 1 H2O + 1 O2 H1 = 226731 + (-3.56 × R × 308.3) = 217605 kJ H2 = 3.5(0 + 6086) = 21301 kJ Products T3 = 350 K = 76.8°C ⇒ PG = 41.8 kPa PG 41.8 nV max yV max = = = 0.0836 = ⇒ nV max = 0.2737 = nV gas mix P 500 nV max + 2 + 1

⇒ nliq = 1 - 0.2737 = 0.7263 Gas Mix = 2 CO2 + 0.2737 H2O + 1 O2 c) Hliq = 0.7263(-285830 + 18.015(321.5 - 104.9)) = -204764 kJ 3

Hgas mix = 2(-393522 + 2036) + 0.2737(-241826 + 1756) + 1541 3

= -847138 kJ H3 = Hliq + Hgas mix = -204 764 -847 138 = -1 051 902 kJ 3

3

H3 - H1 - H2 = -1 290 808 kJ . . . . or H3 - H1 - H2 = -1 290 808/26.038 = -49 574 kW = mH2O(h4 - h5) Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.


h5 = 42.01 +

49574 = 3346.9 15

⇒

T5 = 433.4°C


Fundamentals of Thermodynamics 9th editionch15

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